Section 13-11
Section 13-11 Representations for Cyclic and Related Groups
C +
−
3
C3
1
1
z, x2 + y2, z2
1
1
R2
√
√
−1
3/2
−1 − 3/2
2
√
2
√
(x, y)(R
−
x , Ry )(x2 − y2, xy)(xz, yz)
3/2
−1
3/2
−1
2
2
in e-type MOs. What about 2s on nitrogen? Intuitively, we know that it, like 2pz, is unaffected by all the operations, and so it should appear only in a1 MOs. The table indicates this by listing z2 and x2 + y2 as bases for the A1 representation. Since these have the same symmetry, their sum retains the symmetry, and so x2 + y2 + z2 = r2 is also a basis for A1. Since r2 is spherically symmetric, this indicates that any spherically symmetric function on nitrogen is a basis for the A1 representation.
If we were to use an expanded basis set of AOs, including 3d AOs on nitrogen, the group table tells us that 3dz2 would appear in the nondegenerate a1 MOs and that 3dx2−y2, 3dxy, 3dxz, and 3dyz would appear in degenerate e-type MOs. (Recall that 3dz2 is really 3dz2−r2 or 3d2z2−x2−y2 .)
What about the 1s AOs on the H’s? Can the representation table tell us in which MOs these will appear? It turns out that it can, and that they go into both a1- and e-type MOs, but we will defer showing how this can be told from the table until later in the chapter.
Readers may begin to appreciate the usefulness of group theory in quantum chem istry when they consider that, simply by assigning ammonia to the C3v group and looking up the representation table, we are able to say that a minimum-valence basis set MO calculation will produce nondegenerate, totally symmetric (a1) MOs containing N2s, N2p , and H , N ,
z
1s AOs, and doubly degenerate (e) MOs containing N2px
2py
and H1s AOs. (No a2 MOs will appear because none of our AOs are a basis for that representation.)
13-11 Representations for Cyclic and Related Groups Cyclic groups are the groups C2, C3, C4, . . . , Cn containing only the n − 1 rotation operations and the identity operation E. We devote a separate section to these because there are some special problems connected with finding and labeling representations for these groups. (This section is off the mainstream of development of this chapter and may be skipped if desired.)
Let us consider the operations associated with the n-fold proper axis oriented along the z axis and ask what will become of the function f = exp(iφ) as it is rotated clockwise about this axis by 2π/n radians. (We entertain the idea that exp(iφ) might be a convenient basis for a representation since such functions were found to be eigenfunctions for the particle-in-a-ring problem in Chapter 2.) Since the clockwise direction is opposite to the normal direction of the φ coordinate, the effect of the rotation is to put f (φ) where f (φ − 2π/n) used to be. To see how the function after rotation compares to that before rotation, we must compare exp(iφ) with exp[i(φ − 2π/n)]. That is, the representation Rf , such that C+ n exp(iφ) = Rf exp(˙ιφ), is given by exp(iφ)/ exp[i(φ − 2π/n)], or Chapter 13 Group TheoryTABLE 13-14a Partial Representation for the C3 Group
C3
E
C3
C23
A
1
1
1
z, Rz
1
exp(2π i/3) exp(4π i/3) exp(iφ) TABLE 13-14b Partial Representation for the C3 Group
C3
E
C3
C2 = exp(2πi/3)
3
A
1
1
1
z, Rz
1
∗
exp(iφ) exp(2π i/n), which equals cos(2π/n) + i sin(2π/n). For various fractions of a cycle (i.e., various n), Rf takes on different values: n : 1
2
3
4
5
6
7
8
Rf : 1 −1 exp(2πi/3) i exp(2πi/5) exp(πi/3) exp(2πi/7) exp(πi/4) The important point here is that f = exp(iφ) is a basis for a one-dimensional representation for Cn since a rotation turns f into a constant times f and not into some other function. For the C3 group, then, we could write a partial representation table as shown in Table 13-14a. Now exp(4π i/3) is equal to exp(−2πi/3), and exp(−2πi/3) is the complex conjugate of exp(2π i/3). If we let ≡ exp(2πi/3), we can write the table as shown in Table 13-14b. If f = exp(iφ) is a satisfactory basis, f ∗ = exp(−iφ) is also acceptable, since it is linearly independent of f . If one calculates the representations for n = 1, 2, 3, . . . as before, one finds that the numbers Rf ∗ are the complex conjugates of Rf . Therefore, we can immediately expand our table as shown in Table 13-14c.
Since the existence of exp(iφ) as a basis for a one-dimensional representation always implies that exp(−iφ) exists as a basis, these sorts of one-dimensional representations always occur in pairs. It is conventional to combine these with braces and refer to them with the symbol E, which we claimed earlier is reserved for two-dimensional representations. Note, however, that a pair of one-dimensional representations is not
the same as a two-dimensional representation, and we must broaden our definition of the symbol E to include both types of situation.
Our representation table for the C3 group now looks almost the way one would find it in a standard tabulation. However, instead of listing exp(iφ) and exp(−iφ) as bases, TABLE 13-14c Representation for the C3 Group
C
2
3
E
C3
C3 = exp(2πi/3)
A
1
1
1
z, Rz
E
1
∗
exp(iφ)
1
∗
exp(−iφ)
Section 13-11 Representations for Cyclic and Related GroupsTABLE 13-15 Representation for the C3 Group
C3
E
C3
C23
A
1
1
1
z, Rz
√
√
1
0
−1
3
−1 − 3
E
2
2
√
2
2
√
(x, y)
0
1
− 3
1
3
−1
2
2
2
2
the convention is to list x and y. In relating Cartesian to spherical polar coordinates, x = r sin θ cos φ and y = r sin θ sin φ. When we are concerned only with changes in φ, x goes as cos φ, y as sin φ, and, since sin φ and cos φ are expressible as linear combinations of exp(iφ) and exp(−iφ), it follows that x and y are equivalent to exp(±iφ) as bases. If we had started out with x and y as bases, we would have found that, for rotations, these are intermixed, leading to a truly two-dimensional representation. In fact, we worked out the effects of C3 and C2 on x and y earlier for the C
3
3v group. There we found that C3 and C2(= C−) were represented by the two-dimensional matrices
3
3
shown in Table 13-15. But this E representation is reducible to the two one-dimensional representations through the unitary matrix
∪ = 1
√
1
i
2
i
1
The resulting block-diagonalized representation is
E
C3
C23
√
√
1
0
−1 − i 3
0
−1 + i 3
0
2
2
√
2
2
√
0
1
0
−1 + i 3
0
−1 − i 3
2
2
2
2
which can be separated into two one-dimensional representations:
E
C3
C23
√
√
1
−1 − i 3 −1 + i 3
2
2
2
2
√
√
1
−1 + i 3 −1 − i 3
2
2
2
2
√
Since = exp(2πi/3) = cos(2π/3) + i sin(2π/3) = − 1 + i 3 , we recognize that this
2
2
pair of one-dimensional representations is the same as the pair we found earlier. Furthermore, we note that ∪ is precisely the transformation that turns x (i.e., cos φ) and y (i.e., sin φ) back into exp(±iφ) (to within a constant multiplier):
√
∪ x = 1
√
1
i
cos φ = 1
√
cos φ + i sin φ = (1/ 2) exp(iφ)
√
y
2
i
1
sin φ
2
i cos φ + sin φ (−i/ 2) exp(−iφ) Thus, (x, y) produce a reducible, two-dimensional representation equivalent to the irreducible representation given by exp(±iφ). The reason for listing (x, y) as bases Chapter 13 Group TheoryTABLE 13-16 Irreducible Representations for the C3 Group
C3
E
C3
C2 = exp(2π i/3)
3
A
1
1
1
z, Rz z2, x2 + y2, x2 − y2, xy
1
∗
(x, y)
E
1
∗
(Rx, Ry) (yz, xz) is simply that most applications of the table are made to real functions (e.g., chemists usually prefer to work with 2px and 2py AOs rather than with 2p+1 and 2p−1). Our final form for the table, then, is that shown in Table 13-16, where additional bases have been listed.
The preceding discussion has been within the context of the Cn groups. One might ask what sort of symmetry operation is needed to make it impossible for exp(iφ) to be a basis for a one-dimensional representation. The answer is, any operation that causes a reversal in the direction of the coordinate φ. For then, exp(iφ) → exp(−iφ), and our basis function has turned into another independent function rather than into a constant times itself. Therefore, the presence of any symmetry operation that reverses the direction of motion of the hands of a clock will suffice to prevent representations of the , ∗ sort. Operations that reverse clock direction are σd, σv, and C (perpendicular
2
to the principal axis). Clock direction is unaffected by σh, i, and Sn. Therefore, we can expect , ∗ types of representations to occur in groups of types Cn, Cnh, Sn, all of which have a Cn axis but no σd, σv, or C elements.
2
13-12 Orthogonality in Irreducible InequivalentRepresentations
We come now to a very important point regarding representations. We will illustrate our arguments with the representation table (Table 13-13) for the C3v group. Notice the following features of that table:
1. If we choose the A1 representation, square all the numbers, and sum over all six symmetry operations, we get 6 as a result.
2. If we do the same thing with the A2 representation, we get the same result.
3. If we do the same thing for the upper left-hand elements (the 1, 1 elements) of the E representation, we get 3 as a result.
4. If we do the same thing for each of the other positions in the E representation, the result is 3 each time.
In general, the result of this procedure for any irreducible representation in anygroup will be the order of the group divided by the dimension of the representation. That (j,k) is, if h is the order of the group, li is the dimension of representation i, and
(R)
i
Section 13-12 Orthogonality in Irreducible Inequivalent Representations is the number in the (j, k) position of the matrix representing symmetry operation R, then the mathematical formula that corresponds to our general statement is
2
(j,k)
(R) = h
(13-2)
i
li
R
Note that the absolute square is used to accommodate the , ∗ type representations of cyclic groups.
We can conceive of the set of six numbers for A1 as being a vector of six elements.
The six numbers of A2 constitute a second vector, and E provides four more six √
dimensional vectors. If each such vector is multiplied by li/ h, then each vector is normalized.
Now let us examine these vectors regarding their orthogonality. If we take the scalar product of the unnormalized vectors A1 and A2, the result is zero:
1
−
1
−
1
1 1 1 1 1 1 = 0
−1
A
1
1
1
A2
The reader may quickly verify that the scalar product of any two different vectors from among the set of six in the C3v representation table is zero. Thus these six vectors are orthogonal. Once again, this result always holds between “representation vectors” in irreducible inequivalent representations for any group. Combining this orthogonality property with the normality property mentioned earlier, we have
∗
li/ h(k,l)(R)
l (R) = δ
i
i / h(m,n)
j i,j δk,mδl,n
(13-3)
R where i and j are understood to be irreducible and, if i = j , inequivalent. This relation, sometimes called “the great orthogonality theorem,” is of central importance in group theory. Its essence is captured by the statement that “irreducible inequivalent representations are comprised of orthogonal vectors.” We do not prove the theorem in this book,5 but we do make considerable use of Eq. (13-3).
One immediate result of the relation is that it enables us to tell when we have completed the task of finding all the inequivalent irreducible representations of a group.
If we consider the C3v group, for example, we note that it is of order six, since there are six symmetry operations. This means that each “representation vector” will have six elements, i.e., is a vector in six-dimensional space. The maximum number of orthogonal vectors we can have in six-dimensional space is six. Therefore, the number of representation vectors cannot exceed the order of the group. Furthermore, since the number of vectors provided by an n-dimensional representation is n2 (e.g., E is two-dimensional and gives four vectors), we can state that the sum of the squares of the 5See Bishop [1] or Eyring et al. [2, Appendix VI].
Chapter 13 Group Theorydimensions of the inequivalent irreducible representations of a group cannot exceed
the order of the group.
In fact, it can be proved5 that this sum of squares of dimensions must equal the order of the group when all such representations are included. That is, all inequivalent irreproducible representations
l2 =
i
h
(13-4)
i
Thus, the fact that the squares of the dimensions of the A1, A2, and E representations for the C3v group add up to six, which is the order of the group, indicates that no more irreducible representations exist (except those that are equivalent to those we already have).
EXAMPLE 13-6 Count up all the symmetry operations you can think of for (planar) BH3. Can you guess from this how many irreducible representations exist for this molecule and what their dimensions are?
SOLUTION The operations are: E, C+, C−, σ , S−, C , C or 12 opera 3
3
1, σ2, σ3, σh, S+
3
3
2, C2
2
tions, so the group order is 12. There must be at least one irreducible representation of order one, so we cannot have three two-dimensional representations. We could have: 12 one-dimensional representations; eight one-dimensional and one two-dimensional representations; four one-dimensional and two two-dimensional representations. From what has been presented so far, all three of these are possible.
13-13 Characters and Character Tables
Thus far, we have defined representations and shown how they may be generated from basis functions. We have distinguished between reducible and irreducible representations and have indicated that there is an unlimited number of equivalent representations corresponding to any given two- or higher-dimensional representation. An example of a pair of equivalent, reducible, two-dimensional representations, derived in Section 13-11, is given in Table 13-17. Equivalent representations are related through unitary transformations, which are a special kind of similarity transformation (see Chapter 9), and two matrices that differ only by a similarity transformation have the same TABLE 13-17 Equivalent Representations for the C3 Group
C3
E
C3
C23
√
√
1 0 −1 − 3/2
−1
3/2
2
√
2
√
x,y 0 1
3/2
−1
− 3/2 −1
2
2
√
√
1 0 −1 − i 3/2
0
−1 + i 3/2
0
2
√
2
√
exp(±φ) 0 1
0
−1 + i 3/2
0
−1 − i 3/2
2
2
Section 13-13 Characters and Character TablesTABLE 13-18 Characters for the C3v Group c
+
−
3v
E
σ1
σ2
σ3
C3
C3
A1
1
1
1
1
1
1
z, x2 + y2, z2
A2
1
−1
−1
−1
1
1
Rz
E
2
0
0
0
−1
−1
(x, y)(Rx, Ry) (x2 − y2, xy)(xz, yz) trace, or character (Problem 9-7), which is defined as the sum of the diagonal elements of a matrix. The matrices in Table 13-17 exemplify this fact, their characters being respectively 2, −1, −1 for E, C3, and C2 in both representations. The generally
3
accepted symbol for the character of operation R in representation i is χi(R), and the mathematical definition is j,j χi(R) =
(R)
(13-5)
i
j
In practice, it is the characters of irreducible representations that are used in most chemical applications of group theory. This means that one needs only the character
table for a group, rather than the whole representation table. For the C3v group, the character table is displayed in Table 13-18. Comparison with Table 13-13 will make clear that the character is merely the sum of diagonal elements. (For one-dimensional representations, the character and the representation are identical.) Since the representation for the identity operation is always a unit matrix, the character for this operation is always the same as the dimension of the representation. Hence, the first character in a row tells us the dimension of the corresponding representation.
An immediate consequence of the orthogonality theorem for representations is that the vectors resulting from characters are orthogonal too. This is trivially obvious for characters of one-dimensional representations. For multidimensional representations, the character vector is simply the sum of the representation vectors in diagonal positions.
If a given vector (say, the A1 vector) is orthogonal to each of these (say, E11 and E22), then it is orthogonal to their sum; that is, in terms of the vector notation of Chapter 9, if ˜a1e11 = 0 and ˜a1e22 = 0, then ˜a1(e11 + e22) = 0.
Inspection of Table 13-18 reveals a curious thing. For any given row of characters, all operations in the same class have the same character. There is a fairly simple reason for this. We have indicated already that operations in the same class are operations that can be interchanged merely by group reflections, rotations, etc., of the symmetry elements in the group, but we have seen that such changes are mathematically effected through similarity transformations. This means that representations for operations in the same class are interchangeable via similarity transformations. That is, the matrix representing, say, σ1 (in the E representation of C3v) can be made equal to the matrix representing σ2 through a similarity transformation:
T−1σ1T = σ2
(13-6)
(From our group Table 13-7, we can ascertain that T must be the matrix representing σ3.) Now, the two sides of Eq. (13-6) must have the same character since they are identical 2 × 2 matrices, but the left-hand side must have the same character as σ1 Chapter 13 Group TheoryTABLE 13-19 The Standard Short-Form Character Table for the C3v Group
C3v
E
3σ
2C3
A1
1
1
1
z, x2 + y2, z2
A2
1
−1
1
Rz
E
2
0
−1
(x, y)(Rx, Ry)(x2 − y2, xy)(xz, yz) since character is unchanged by a similarity transformation. Therefore, σ1 and σ2 have the same character.
We can take advantage of the above rule to write our character table in abbreviated form, illustrated for the C3v group in Table 13-19. This is the standard form for character tables. A collection of such tables appears in Appendix 11.
That characters must be equal in the same class is a restriction on our character vectors. In Table 13-19 it is made evident that, in the C3v group, our vectors really only have three independent variables each, one for each class. These are properly thought of, then, as vectors in three-dimensional space (with weighting factors 1, 3, and 2 for E, σ , and C3, respectively). There can be no more than three such vectors that are orthogonal, and so we are left with the result that the number of inequivalent irreducible representations in a group cannot exceed (and is in fact equal to6) the number of classes in the group. This result, together with the fact that the sum of squares of dimensions of inequivalent irreducible representations must equal the order of the group, often suffices to tell us in advance how many representations there are and what their dimensions are. For our C3v group, the order is six and there are three classes. Hence, we know that there are three representations and that the squares of their dimensions sum to six.
The problem reduces to: “What three positive integers squared, sum to six?” There is only one answer: 1, 1, and 2. The fact that there will always be a totally symmetric one-dimensional representation also helps pin down the possibilities. For example, can one have a group of order eight and only two classes? There is no way this can happen.
Two classes would mean two representations. If both were E type, their dimensions squared would indeed sum to eight. But one of them must be one-dimensional, and there is no way the other can square to seven.
EXAMPLE 13-7 How many classes of symmetry operation can you count for (pla nar) BH3? Can you use this to select one of the possibilities from Example 13-12?
SOLUTION The classes are E, σv, C3, C2, σh, S3. That’s six classes, so there are only six irreducible representations. There are four one-dimensional and two two-dimensional representations.
Our collected list of conditions that the characters in a completed table must satisfy is as follows:
1. There must be a one-dimensional representation having all characters equal to +1.
2. The leading character in each row (i.e., the character for operation E) must equal the dimension of the representation.
6See Bishop [1].
Section 13-13 Characters and Character Tables3. The sum of the squares of the leading characters must equal the order of the group.
4. The number of rows in the character table must equal the number of classes in the group.
5. The absolute squares of the characters in a given row (times the weighting factor for each class if the abbreviated form is used) equals the order of the group. (Character vectors are normalized) This results directly from Eq. (13-2).
6. The character vectors are orthogonal (again, using weighting factors, if appropriate).
This is a fairly large number of restrictions, and may suffice to allow one to produce the character table for a group without ever actually producing representations. For example, consider the C4v group, which has the operations E, 2C4, C2, 2σv, and 2σd.
The group thus has order eight and five classes. There must be five representations, and the only way their dimensions can square to eight is if four of them are one-dimensional and one is two-dimensional. This already enables us to write Table 13-20. It is not TABLE 13-20 Partial C4v Character Table
C4v
E
2C4
C2
2σv
2σd
A1
1
1
1
1
1
1
1
2
1
3
1
E
2
difficult to find a way to make 1 orthogonal to A1. We simply place −1 in some places to produce four products of −1 and four of +1. Three possibilities are shown in Table 13-21. These are orthogonal not only to A1, but to each other as well, and so we TABLE 13-21 Partial C4v Character Table
E
2C4
C2
2σv
2σd
1
1
1
−1
−1
1
−1
1
1
−1
1
−1
1
−1
1
have found the characters for 1, 2, and 3. The characters for the E representation must have squares that sum to eight and also be orthogonal to all four one-dimensional representation vectors. One possibility is fairly obvious. Since the characters for operations E and C2 are +1 in all the one-dimensional representations, we could take the characters for the E representation to be
E
2C4
C2
2σv
2σd
2
0
−2
0
0
Chapter 13 Group Theory
Other cases that meet the normality condition are
2
±1 0 ±1
0
2
0
0
±1 ±1
2
±1 0
0
±1
But none of these is orthogonal to all the one-dimensional sets. Therefore, the complete character table (except for the basis functions) for the C4v group is shown in Table 13-22, where the symbols A2, B1, B2 are consistent with symmetry or antisymmetry for C4 and σv as described in Section 13-9.
TABLE 13-22 Completed C4v Character Table
C4v
E
2C4
C2
2σv
2σd
A1
1
1
1
1
1
A2
1
1
1
−1
−1
B1
1
−1
1
1
−1
B2
1
−1
1
−1
1
E
2
0
−2
0
0
13-14 Using Characters to Resolve ReducibleRepresentations
It was pointed out earlier that several irreducible representations can be combined into a larger-dimensional reducible representation. Our example was
A 2
0
0
2
0
0
0
1
0 ≡ 0
A1
0 =
0
0
3
0
0
E which is a four-dimensional representation (since 3 is two-dimensional). A matrix built up in this way is symbolized A2 ⊕ A1 ⊕ E. It is easy to see that the characters of the reducible representation are simply the sums of characters for the individual irreducible component representations (since the diagonal elements of A2, A1, and E all lie on the diagonal of ). Thus, the characters of are
E
3σ
2C3
:
4
0
1
Furthermore, no matter how is disguised by a similarity transformation, its character vector is unchanged. Now, suppose you were given the representation , disguised through some similarity transformation so as to be nonblock diagonal and asked to tell which irreducible representations were present. How could you do it? One way would be to find the similarity transformation that would return the representation to block
C +
−
3
C3
1
1
z, x2 + y2, z2
1
1
R2
√
√
−1
3/2
−1 − 3/2
2
√
2
√
(x, y)(R
−
x , Ry )(x2 − y2, xy)(xz, yz)
3/2
−1
3/2
−1
2
2
in e-type MOs. What about 2s on nitrogen? Intuitively, we know that it, like 2pz, is unaffected by all the operations, and so it should appear only in a1 MOs. The table indicates this by listing z2 and x2 + y2 as bases for the A1 representation. Since these have the same symmetry, their sum retains the symmetry, and so x2 + y2 + z2 = r2 is also a basis for A1. Since r2 is spherically symmetric, this indicates that any spherically symmetric function on nitrogen is a basis for the A1 representation.
If we were to use an expanded basis set of AOs, including 3d AOs on nitrogen, the group table tells us that 3dz2 would appear in the nondegenerate a1 MOs and that 3dx2−y2, 3dxy, 3dxz, and 3dyz would appear in degenerate e-type MOs. (Recall that 3dz2 is really 3dz2−r2 or 3d2z2−x2−y2 .)
What about the 1s AOs on the H’s? Can the representation table tell us in which MOs these will appear? It turns out that it can, and that they go into both a1- and e-type MOs, but we will defer showing how this can be told from the table until later in the chapter.
Readers may begin to appreciate the usefulness of group theory in quantum chem istry when they consider that, simply by assigning ammonia to the C3v group and looking up the representation table, we are able to say that a minimum-valence basis set MO calculation will produce nondegenerate, totally symmetric (a1) MOs containing N2s, N2p , and H , N ,
z
1s AOs, and doubly degenerate (e) MOs containing N2px
2py
and H1s AOs. (No a2 MOs will appear because none of our AOs are a basis for that representation.)
13-11 Representations for Cyclic and Related Groups Cyclic groups are the groups C2, C3, C4, . . . , Cn containing only the n − 1 rotation operations and the identity operation E. We devote a separate section to these because there are some special problems connected with finding and labeling representations for these groups. (This section is off the mainstream of development of this chapter and may be skipped if desired.)
Let us consider the operations associated with the n-fold proper axis oriented along the z axis and ask what will become of the function f = exp(iφ) as it is rotated clockwise about this axis by 2π/n radians. (We entertain the idea that exp(iφ) might be a convenient basis for a representation since such functions were found to be eigenfunctions for the particle-in-a-ring problem in Chapter 2.) Since the clockwise direction is opposite to the normal direction of the φ coordinate, the effect of the rotation is to put f (φ) where f (φ − 2π/n) used to be. To see how the function after rotation compares to that before rotation, we must compare exp(iφ) with exp[i(φ − 2π/n)]. That is, the representation Rf , such that C+ n exp(iφ) = Rf exp(˙ιφ), is given by exp(iφ)/ exp[i(φ − 2π/n)], or Chapter 13 Group TheoryTABLE 13-14a Partial Representation for the C3 Group
C3
E
C3
C23
A
1
1
1
z, Rz
1
exp(2π i/3) exp(4π i/3) exp(iφ) TABLE 13-14b Partial Representation for the C3 Group
C3
E
C3
C2 = exp(2πi/3)
3
A
1
1
1
z, Rz
1
∗
exp(iφ) exp(2π i/n), which equals cos(2π/n) + i sin(2π/n). For various fractions of a cycle (i.e., various n), Rf takes on different values: n : 1
2
3
4
5
6
7
8
Rf : 1 −1 exp(2πi/3) i exp(2πi/5) exp(πi/3) exp(2πi/7) exp(πi/4) The important point here is that f = exp(iφ) is a basis for a one-dimensional representation for Cn since a rotation turns f into a constant times f and not into some other function. For the C3 group, then, we could write a partial representation table as shown in Table 13-14a. Now exp(4π i/3) is equal to exp(−2πi/3), and exp(−2πi/3) is the complex conjugate of exp(2π i/3). If we let ≡ exp(2πi/3), we can write the table as shown in Table 13-14b. If f = exp(iφ) is a satisfactory basis, f ∗ = exp(−iφ) is also acceptable, since it is linearly independent of f . If one calculates the representations for n = 1, 2, 3, . . . as before, one finds that the numbers Rf ∗ are the complex conjugates of Rf . Therefore, we can immediately expand our table as shown in Table 13-14c.
Since the existence of exp(iφ) as a basis for a one-dimensional representation always implies that exp(−iφ) exists as a basis, these sorts of one-dimensional representations always occur in pairs. It is conventional to combine these with braces and refer to them with the symbol E, which we claimed earlier is reserved for two-dimensional representations. Note, however, that a pair of one-dimensional representations is not
the same as a two-dimensional representation, and we must broaden our definition of the symbol E to include both types of situation.
Our representation table for the C3 group now looks almost the way one would find it in a standard tabulation. However, instead of listing exp(iφ) and exp(−iφ) as bases, TABLE 13-14c Representation for the C3 Group
C
2
3
E
C3
C3 = exp(2πi/3)
A
1
1
1
z, Rz
E
1
∗
exp(iφ)
1
∗
exp(−iφ)
Section 13-11 Representations for Cyclic and Related GroupsTABLE 13-15 Representation for the C3 Group
C3
E
C3
C23
A
1
1
1
z, Rz
√
√
1
0
−1
3
−1 − 3
E
2
2
√
2
2
√
(x, y)
0
1
− 3
1
3
−1
2
2
2
2
the convention is to list x and y. In relating Cartesian to spherical polar coordinates, x = r sin θ cos φ and y = r sin θ sin φ. When we are concerned only with changes in φ, x goes as cos φ, y as sin φ, and, since sin φ and cos φ are expressible as linear combinations of exp(iφ) and exp(−iφ), it follows that x and y are equivalent to exp(±iφ) as bases. If we had started out with x and y as bases, we would have found that, for rotations, these are intermixed, leading to a truly two-dimensional representation. In fact, we worked out the effects of C3 and C2 on x and y earlier for the C
3
3v group. There we found that C3 and C2(= C−) were represented by the two-dimensional matrices
3
3
shown in Table 13-15. But this E representation is reducible to the two one-dimensional representations through the unitary matrix
∪ = 1
√
1
i
2
i
1
The resulting block-diagonalized representation is
E
C3
C23
√
√
1
0
−1 − i 3
0
−1 + i 3
0
2
2
√
2
2
√
0
1
0
−1 + i 3
0
−1 − i 3
2
2
2
2
which can be separated into two one-dimensional representations:
E
C3
C23
√
√
1
−1 − i 3 −1 + i 3
2
2
2
2
√
√
1
−1 + i 3 −1 − i 3
2
2
2
2
√
Since = exp(2πi/3) = cos(2π/3) + i sin(2π/3) = − 1 + i 3 , we recognize that this
2
2
pair of one-dimensional representations is the same as the pair we found earlier. Furthermore, we note that ∪ is precisely the transformation that turns x (i.e., cos φ) and y (i.e., sin φ) back into exp(±iφ) (to within a constant multiplier):
√
∪ x = 1
√
1
i
cos φ = 1
√
cos φ + i sin φ = (1/ 2) exp(iφ)
√
y
2
i
1
sin φ
2
i cos φ + sin φ (−i/ 2) exp(−iφ) Thus, (x, y) produce a reducible, two-dimensional representation equivalent to the irreducible representation given by exp(±iφ). The reason for listing (x, y) as bases Chapter 13 Group TheoryTABLE 13-16 Irreducible Representations for the C3 Group
C3
E
C3
C2 = exp(2π i/3)
3
A
1
1
1
z, Rz z2, x2 + y2, x2 − y2, xy
1
∗
(x, y)
E
1
∗
(Rx, Ry) (yz, xz) is simply that most applications of the table are made to real functions (e.g., chemists usually prefer to work with 2px and 2py AOs rather than with 2p+1 and 2p−1). Our final form for the table, then, is that shown in Table 13-16, where additional bases have been listed.
The preceding discussion has been within the context of the Cn groups. One might ask what sort of symmetry operation is needed to make it impossible for exp(iφ) to be a basis for a one-dimensional representation. The answer is, any operation that causes a reversal in the direction of the coordinate φ. For then, exp(iφ) → exp(−iφ), and our basis function has turned into another independent function rather than into a constant times itself. Therefore, the presence of any symmetry operation that reverses the direction of motion of the hands of a clock will suffice to prevent representations of the , ∗ sort. Operations that reverse clock direction are σd, σv, and C (perpendicular
2
to the principal axis). Clock direction is unaffected by σh, i, and Sn. Therefore, we can expect , ∗ types of representations to occur in groups of types Cn, Cnh, Sn, all of which have a Cn axis but no σd, σv, or C elements.
2
13-12 Orthogonality in Irreducible InequivalentRepresentations
We come now to a very important point regarding representations. We will illustrate our arguments with the representation table (Table 13-13) for the C3v group. Notice the following features of that table:
1. If we choose the A1 representation, square all the numbers, and sum over all six symmetry operations, we get 6 as a result.
2. If we do the same thing with the A2 representation, we get the same result.
3. If we do the same thing for the upper left-hand elements (the 1, 1 elements) of the E representation, we get 3 as a result.
4. If we do the same thing for each of the other positions in the E representation, the result is 3 each time.
In general, the result of this procedure for any irreducible representation in anygroup will be the order of the group divided by the dimension of the representation. That (j,k) is, if h is the order of the group, li is the dimension of representation i, and
(R)
i
Section 13-12 Orthogonality in Irreducible Inequivalent Representations is the number in the (j, k) position of the matrix representing symmetry operation R, then the mathematical formula that corresponds to our general statement is
2
(j,k)
(R) = h
(13-2)
i
li
R
Note that the absolute square is used to accommodate the , ∗ type representations of cyclic groups.
We can conceive of the set of six numbers for A1 as being a vector of six elements.
The six numbers of A2 constitute a second vector, and E provides four more six √
dimensional vectors. If each such vector is multiplied by li/ h, then each vector is normalized.
Now let us examine these vectors regarding their orthogonality. If we take the scalar product of the unnormalized vectors A1 and A2, the result is zero:
1
−
1
−
1
1 1 1 1 1 1 = 0
−1
A
1
1
1
A2
The reader may quickly verify that the scalar product of any two different vectors from among the set of six in the C3v representation table is zero. Thus these six vectors are orthogonal. Once again, this result always holds between “representation vectors” in irreducible inequivalent representations for any group. Combining this orthogonality property with the normality property mentioned earlier, we have
∗
li/ h(k,l)(R)
l (R) = δ
i
i / h(m,n)
j i,j δk,mδl,n
(13-3)
R where i and j are understood to be irreducible and, if i = j , inequivalent. This relation, sometimes called “the great orthogonality theorem,” is of central importance in group theory. Its essence is captured by the statement that “irreducible inequivalent representations are comprised of orthogonal vectors.” We do not prove the theorem in this book,5 but we do make considerable use of Eq. (13-3).
One immediate result of the relation is that it enables us to tell when we have completed the task of finding all the inequivalent irreducible representations of a group.
If we consider the C3v group, for example, we note that it is of order six, since there are six symmetry operations. This means that each “representation vector” will have six elements, i.e., is a vector in six-dimensional space. The maximum number of orthogonal vectors we can have in six-dimensional space is six. Therefore, the number of representation vectors cannot exceed the order of the group. Furthermore, since the number of vectors provided by an n-dimensional representation is n2 (e.g., E is two-dimensional and gives four vectors), we can state that the sum of the squares of the 5See Bishop [1] or Eyring et al. [2, Appendix VI].
Chapter 13 Group Theorydimensions of the inequivalent irreducible representations of a group cannot exceed
the order of the group.
In fact, it can be proved5 that this sum of squares of dimensions must equal the order of the group when all such representations are included. That is, all inequivalent irreproducible representations
l2 =
i
h
(13-4)
i
Thus, the fact that the squares of the dimensions of the A1, A2, and E representations for the C3v group add up to six, which is the order of the group, indicates that no more irreducible representations exist (except those that are equivalent to those we already have).
EXAMPLE 13-6 Count up all the symmetry operations you can think of for (planar) BH3. Can you guess from this how many irreducible representations exist for this molecule and what their dimensions are?
SOLUTION The operations are: E, C+, C−, σ , S−, C , C or 12 opera 3
3
1, σ2, σ3, σh, S+
3
3
2, C2
2
tions, so the group order is 12. There must be at least one irreducible representation of order one, so we cannot have three two-dimensional representations. We could have: 12 one-dimensional representations; eight one-dimensional and one two-dimensional representations; four one-dimensional and two two-dimensional representations. From what has been presented so far, all three of these are possible.
13-13 Characters and Character Tables
Thus far, we have defined representations and shown how they may be generated from basis functions. We have distinguished between reducible and irreducible representations and have indicated that there is an unlimited number of equivalent representations corresponding to any given two- or higher-dimensional representation. An example of a pair of equivalent, reducible, two-dimensional representations, derived in Section 13-11, is given in Table 13-17. Equivalent representations are related through unitary transformations, which are a special kind of similarity transformation (see Chapter 9), and two matrices that differ only by a similarity transformation have the same TABLE 13-17 Equivalent Representations for the C3 Group
C3
E
C3
C23
√
√
1 0 −1 − 3/2
−1
3/2
2
√
2
√
x,y 0 1
3/2
−1
− 3/2 −1
2
2
√
√
1 0 −1 − i 3/2
0
−1 + i 3/2
0
2
√
2
√
exp(±φ) 0 1
0
−1 + i 3/2
0
−1 − i 3/2
2
2
Section 13-13 Characters and Character TablesTABLE 13-18 Characters for the C3v Group c
+
−
3v
E
σ1
σ2
σ3
C3
C3
A1
1
1
1
1
1
1
z, x2 + y2, z2
A2
1
−1
−1
−1
1
1
Rz
E
2
0
0
0
−1
−1
(x, y)(Rx, Ry) (x2 − y2, xy)(xz, yz) trace, or character (Problem 9-7), which is defined as the sum of the diagonal elements of a matrix. The matrices in Table 13-17 exemplify this fact, their characters being respectively 2, −1, −1 for E, C3, and C2 in both representations. The generally
3
accepted symbol for the character of operation R in representation i is χi(R), and the mathematical definition is j,j χi(R) =
(R)
(13-5)
i
j
In practice, it is the characters of irreducible representations that are used in most chemical applications of group theory. This means that one needs only the character
table for a group, rather than the whole representation table. For the C3v group, the character table is displayed in Table 13-18. Comparison with Table 13-13 will make clear that the character is merely the sum of diagonal elements. (For one-dimensional representations, the character and the representation are identical.) Since the representation for the identity operation is always a unit matrix, the character for this operation is always the same as the dimension of the representation. Hence, the first character in a row tells us the dimension of the corresponding representation.
An immediate consequence of the orthogonality theorem for representations is that the vectors resulting from characters are orthogonal too. This is trivially obvious for characters of one-dimensional representations. For multidimensional representations, the character vector is simply the sum of the representation vectors in diagonal positions.
If a given vector (say, the A1 vector) is orthogonal to each of these (say, E11 and E22), then it is orthogonal to their sum; that is, in terms of the vector notation of Chapter 9, if ˜a1e11 = 0 and ˜a1e22 = 0, then ˜a1(e11 + e22) = 0.
Inspection of Table 13-18 reveals a curious thing. For any given row of characters, all operations in the same class have the same character. There is a fairly simple reason for this. We have indicated already that operations in the same class are operations that can be interchanged merely by group reflections, rotations, etc., of the symmetry elements in the group, but we have seen that such changes are mathematically effected through similarity transformations. This means that representations for operations in the same class are interchangeable via similarity transformations. That is, the matrix representing, say, σ1 (in the E representation of C3v) can be made equal to the matrix representing σ2 through a similarity transformation:
T−1σ1T = σ2
(13-6)
(From our group Table 13-7, we can ascertain that T must be the matrix representing σ3.) Now, the two sides of Eq. (13-6) must have the same character since they are identical 2 × 2 matrices, but the left-hand side must have the same character as σ1 Chapter 13 Group TheoryTABLE 13-19 The Standard Short-Form Character Table for the C3v Group
C3v
E
3σ
2C3
A1
1
1
1
z, x2 + y2, z2
A2
1
−1
1
Rz
E
2
0
−1
(x, y)(Rx, Ry)(x2 − y2, xy)(xz, yz) since character is unchanged by a similarity transformation. Therefore, σ1 and σ2 have the same character.
We can take advantage of the above rule to write our character table in abbreviated form, illustrated for the C3v group in Table 13-19. This is the standard form for character tables. A collection of such tables appears in Appendix 11.
That characters must be equal in the same class is a restriction on our character vectors. In Table 13-19 it is made evident that, in the C3v group, our vectors really only have three independent variables each, one for each class. These are properly thought of, then, as vectors in three-dimensional space (with weighting factors 1, 3, and 2 for E, σ , and C3, respectively). There can be no more than three such vectors that are orthogonal, and so we are left with the result that the number of inequivalent irreducible representations in a group cannot exceed (and is in fact equal to6) the number of classes in the group. This result, together with the fact that the sum of squares of dimensions of inequivalent irreducible representations must equal the order of the group, often suffices to tell us in advance how many representations there are and what their dimensions are. For our C3v group, the order is six and there are three classes. Hence, we know that there are three representations and that the squares of their dimensions sum to six.
The problem reduces to: “What three positive integers squared, sum to six?” There is only one answer: 1, 1, and 2. The fact that there will always be a totally symmetric one-dimensional representation also helps pin down the possibilities. For example, can one have a group of order eight and only two classes? There is no way this can happen.
Two classes would mean two representations. If both were E type, their dimensions squared would indeed sum to eight. But one of them must be one-dimensional, and there is no way the other can square to seven.
EXAMPLE 13-7 How many classes of symmetry operation can you count for (pla nar) BH3? Can you use this to select one of the possibilities from Example 13-12?
SOLUTION The classes are E, σv, C3, C2, σh, S3. That’s six classes, so there are only six irreducible representations. There are four one-dimensional and two two-dimensional representations.
Our collected list of conditions that the characters in a completed table must satisfy is as follows:
1. There must be a one-dimensional representation having all characters equal to +1.
2. The leading character in each row (i.e., the character for operation E) must equal the dimension of the representation.
6See Bishop [1].
Section 13-13 Characters and Character Tables3. The sum of the squares of the leading characters must equal the order of the group.
4. The number of rows in the character table must equal the number of classes in the group.
5. The absolute squares of the characters in a given row (times the weighting factor for each class if the abbreviated form is used) equals the order of the group. (Character vectors are normalized) This results directly from Eq. (13-2).
6. The character vectors are orthogonal (again, using weighting factors, if appropriate).
This is a fairly large number of restrictions, and may suffice to allow one to produce the character table for a group without ever actually producing representations. For example, consider the C4v group, which has the operations E, 2C4, C2, 2σv, and 2σd.
The group thus has order eight and five classes. There must be five representations, and the only way their dimensions can square to eight is if four of them are one-dimensional and one is two-dimensional. This already enables us to write Table 13-20. It is not TABLE 13-20 Partial C4v Character Table
C4v
E
2C4
C2
2σv
2σd
A1
1
1
1
1
1
1
1
2
1
3
1
E
2
difficult to find a way to make 1 orthogonal to A1. We simply place −1 in some places to produce four products of −1 and four of +1. Three possibilities are shown in Table 13-21. These are orthogonal not only to A1, but to each other as well, and so we TABLE 13-21 Partial C4v Character Table
E
2C4
C2
2σv
2σd
1
1
1
−1
−1
1
−1
1
1
−1
1
−1
1
−1
1
have found the characters for 1, 2, and 3. The characters for the E representation must have squares that sum to eight and also be orthogonal to all four one-dimensional representation vectors. One possibility is fairly obvious. Since the characters for operations E and C2 are +1 in all the one-dimensional representations, we could take the characters for the E representation to be
E
2C4
C2
2σv
2σd
2
0
−2
0
0
Chapter 13 Group Theory
Other cases that meet the normality condition are
2
±1 0 ±1
0
2
0
0
±1 ±1
2
±1 0
0
±1
But none of these is orthogonal to all the one-dimensional sets. Therefore, the complete character table (except for the basis functions) for the C4v group is shown in Table 13-22, where the symbols A2, B1, B2 are consistent with symmetry or antisymmetry for C4 and σv as described in Section 13-9.
TABLE 13-22 Completed C4v Character Table
C4v
E
2C4
C2
2σv
2σd
A1
1
1
1
1
1
A2
1
1
1
−1
−1
B1
1
−1
1
1
−1
B2
1
−1
1
−1
1
E
2
0
−2
0
0
13-14 Using Characters to Resolve ReducibleRepresentations
It was pointed out earlier that several irreducible representations can be combined into a larger-dimensional reducible representation. Our example was
A 2
0
0
2
0
0
0
1
0 ≡ 0
A1
0 =
0
0
3
0
0
E which is a four-dimensional representation (since 3 is two-dimensional). A matrix built up in this way is symbolized A2 ⊕ A1 ⊕ E. It is easy to see that the characters of the reducible representation are simply the sums of characters for the individual irreducible component representations (since the diagonal elements of A2, A1, and E all lie on the diagonal of ). Thus, the characters of are
E
3σ
2C3
:
4
0
1
Furthermore, no matter how is disguised by a similarity transformation, its character vector is unchanged. Now, suppose you were given the representation , disguised through some similarity transformation so as to be nonblock diagonal and asked to tell which irreducible representations were present. How could you do it? One way would be to find the similarity transformation that would return the representation to block