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926 C. GROUP THEORY IN SPECTROSCOPY If we obtain another orbital (ϕ 2 ), then we may begin to play with it by applying all the symmetry operations. Some operations will lead to the same (new) orbital, sometimes with the opposite sign. After other operations we will obtain the old orbital ϕ 1 , sometimes with the opposite sign, and sometimes these operations will lead to a third orbital ϕ 3 . Then we apply the symmetry operations to the third or- bital, etc. until the final set of orbitals is obtained which transform into themselves when subject to symmetry operations. The set of such linearly independent orbitals ϕ i , i =1n, may be treated as the basis set in a vector space. All the results of the application of operation ˆ R i on the orbitals ϕ i are collected in a transformation matrix R i : ˆ R i ϕ =R T i ϕ where ϕ = ⎡ ⎣ ϕ 1 ϕ n ⎤ ⎦ (C.16) The matrices R i , i = 1 2g,formthen-dimensional representation (in general reducible) of the symmetry group of the molecule. Indeed, let us see what happens if we apply operation ˆ T = ˆ R 1 ˆ R 2 to the func- tion ϕ i : ˆ T ϕ i = ˆ R 1 ˆ R 2 ϕ i = ˆ R 1 R T 2 ϕ =R T 2 ˆ R 1 ϕ =R T 2 R T 1 ϕ =(R 1 R 2 ) T ϕ This means that all the matrices R i form a representation. BASIS OF A REPRESENTATION The set of linearly independent functions ϕ i , which served to create the representation, forms the basis of the representation. The basis need not have been composed of the orbitals, it could be expressions like x, y, z or x 2 , y 2 , z 2 , xy, xz, yz or any linearly independent functions, provided they transform into themselves under symmetry operations. We may begin from an atomic orbital, and after applying symmetry operations will soon obtain a basis set which contains this orbital and all the other equivalent orbitals. Decomposition of a function into irreducible representation components Letustakeafunctionf belonging to a Hilbert space. Since (see eq. (C.15)) α ˆ P (α) = 1, where α goes over all the irreducible representations of the group, f can be written as the sum of its components f α , each component (belonging to the corresponding subspace of the Hilbert space) transforming according to the irreducible representation α f =1 ·f = α ˆ P (α) f = α f α (C.17) In view of (C.14), components f α and f β are automatically orthogonal if α =β 3 Group theory and quantum mechanics 927 Example 12. Decomposition of a function. Let us take three hydrogen atoms in the configuration of an equilateral triangle, and assume we are in the simplest version of the molecular orbitals in the LCAO MO approximation, i.e. the atomic basis set is composed of the three 1s orbitals ab c centred on the three nuclei. Let us check whether the following functions: u 1 = a +b +c u 2 = b −c u 3 = a −c form the basis to a (reducible) representation. If symmetry operations are applied to ab c, they transform into each other (cf. Fig. C.2), and the results obtained are easily shown as linear combinations of the functions u 1 u 2 u 3 (with R T i as transformation matrices). For example (see Table C.1, p. 911), ˆ Au 1 =a +b +c = u 1 , ˆ Au 2 =−b +c =−u 2 , ˆ Au 3 =a −b =−u 2 +u 3 . Hence, A T = ⎡ ⎣ 100 0 −10 0 −11 ⎤ ⎦ (C.18) In this way (see (C.16)) we obtain R i as: E = ⎡ ⎣ 100 010 001 ⎤ ⎦ A = ⎡ ⎣ 100 0 −1 −1 001 ⎤ ⎦ B = ⎡ ⎣ 100 010 0 −1 −1 ⎤ ⎦ (C.19) C = ⎡ ⎣ 100 001 010 ⎤ ⎦ D = ⎡ ⎣ 100 001 0 −1 −1 ⎤ ⎦ F = ⎡ ⎣ 100 0 −1 −1 010 ⎤ ⎦ (C.20) Let us check that DF = E and AD = B, i.e. exactly as for operations: ˆ D ˆ F = ˆ E, ˆ A ˆ D = ˆ B, and so on. Thus this is a representation, moreover, this is a representa- tion in a block form, because u 1 always transforms within itself, while u 2 and u 3 mix between themselves. It can be shown that this mixing cannot be avoided by any choice of u. Hence, u 1 alone represents the basis of a one-dimensional irre- ducible representation (A 1 – this is seen from the characters corresponding to the first block 1 ×1), while u 2 and u 3 form the basis of a two-dimensional irreducible representation (E). Note that from the mathematical form of the functions u,it follows that u 2 and u 3 have to correspond to the same energy and this energy is different from that corresponding to u 1 . The conclusion is that ab c form the ba- sis for a reducible representation, while their linear combinations u 1 and {u 2 u 3 } form the basis sets of two irreducible representations: A 1 and E. Any function which is a linear combination of a bc can also be represented as a linear combi- nation of u 1 u 2 u 3 . The same symmetry orbitals can be obtained using the projection operators (C.13). Let us take any of functions a b c (the result does not depend on this 928 C. GROUP THEORY IN SPECTROSCOPY choice), e.g., function a. In view of the beautiful equilateral triangle, such a func- tion is no doubt a deformed object, which does not take the trouble to make the three vertices of the triangle equivalent. Let us see whether such a function has any component which transforms according to the irreducible representation A 1 . To this end, let us use the projection operator ˆ P (A 1 ) (Table C.5 of characters on p. 921 and eq. (C.13)): ˆ P (A 1 ) a = 1 6 (a +b +c +a +b +c) = 1 3 (a +b +c).There is thus a fully symmetric component 31 in a. Now, let us use the same orbital a to obtain: ˆ P (A 2 ) a = 1 6 (a +b +c −a −b −c) =0. This means that a does not contain anything which transforms according to A 2 . Now it is the turn of the irreducible representation E: ˆ P (E) a = 2 6 (2a −b−c +0·a +0·b+0·c) = 1 3 [2(a −c)−(b −c)]. We have now obtained is a linear combination of u 2 and u 3 . If the projections were made for function b, we would obtain a trivial repeti- tion 32 for the irreducible representations A 1 and A 2 and a non-trivial result for the irreducible representation E: ˆ P (E) b = 2 6 (2b −a − c + 0 · a + 0 ·b +0 · c) = 1 3 [2(b −c)−(a −c)], which is just another linear combination of u 2 and u 3 .These two functions are therefore inseparable and form the basis for a two-dimensional irreducible representation. DECOMPOSITION INTO IRREDUCIBLE REPRESENTATIONS Any function that is a linear combination of the basis functions of a reducible representation can be decomposed into a linear combination of the basis functions of those irreducible representations which form the reducible rep- resentation. Most important MOST IMPORTANT. . . The wavefunctions corresponding to an energy level – form the basis of an irreducible representation of the symmetry group of the molecule, or in other words, transform according to this irreducible representation – the dimension of the representation is equal to the degeneracy of the en- ergy level. This is how it should be, because if a symmetry operation acts on an eigen- function of the Hamiltonian, we will have only two possible results: 1) we obtain the same function to the accuracy of the sign (in the case of a one-dimensional representation, by definition irreducible); 2) another function corresponding to the 31 This sentence carries a simple message, that by mixing symmetric objects we may obtain an asymmet- ric object, e.g., the asymmetric function a +2b can be represented by the linear combination u 1 +u 2 , both functions transforming according to an irreducible representation of the symmetry group. 32 ˆ P (A 1 ) b = 1 3 (a +b +c) and ˆ P (A 2 ) b =0. 4 Integrals important in spectroscopy 929 Fig. C.4. Each energy level corresponds to an irreducible rep- resentation of the symmetry group of the Hamiltonian. Its linearly independent eigenfunctions corresponding to a given level form the basis of the irreducible representation, or in other words, transform according to this representation. The number of the basis functions is equal to the degeneracy of the level. same energy (because of the same physical situation). Acting on the function ob- tained and repeating the whole procedure, we will finally arrive at a set of n lin- early independent functions which correspond to the same energy (the basis of a n-dimensional irreducible representation). This means (Fig. C.4) that the energy levels may be labelled, each label corresponding to a single irre- ducible representation. Eugene Wigner was the first who obtained this result. This will be of fundamen- tal importance when the selection rules in spectroscopy will be considered. We usually have plenty of energy levels, while the number of irreducible repre- sentations is small. Thus, in general there will be many levels with the same label. Group theory will never tell us either how many levels correspond to a particular irreducible representation, or to what energy they correspond. 4 INTEGRALS IMPORTANT IN SPECTROSCOPY Direct product of the irreducible representations We are quickly approaching the application of group theory to optical transitions in spectroscopy. The most important issue here will be a decision as to whether an integral is zero or non-zero. If the integral is zero, the transition is forbidden, if it is non-zero, it is allowed. To make such a decision we have to use what is known as the direct product of irreducible representations. Imagine basis functions {ϕ i } and {ψ j } which correspond to irreducible representations α and β of the symmetry group of a molecule. Let us make a set {ϕ i ψ j } of all their possible products (i.e. the Cartesian product). 930 C. GROUP THEORY IN SPECTROSCOPY DIRECT PRODUCT The products {ϕ i ψ j }when subject to symmetry operations, lead (as usual) to a representation, and this representation is called the direct product α × β of the irreducible representations α and β . The functions {ϕ i ψ j } form the basis set of a representation (reducible in gen- eral). The matrices of the representations we obtain as usual by applying symmetry operations (eq. (C.16)): ˆ R ϕ i (r)ψ j (r) = ϕ i ˆ R −1 r ψ j ˆ R −1 r = k (α) ki ϕ k l (β) lj ψ l = kl (α) ki (β) lj ϕ k ψ l = kl Z ijkl ϕ k ψ l where (γ) ki are the matrix elements of the irreducible representation γ Z ijkl = (α) ki (β) lj . Of course, the dimension of this representation is the product of the dimensions of the representations α and β, because this is the number of the functions ϕ k ψ l . The characters χ (α×β) of the representation can easily be obtained from the characters of the irreducible representations, we just have to multiply the latter: χ (α×β) ˆ R =χ (α) ˆ R χ (β) ˆ R (C.21) Indeed, the formula is justified by: χ (α×β) ˆ R = kl Z klkl = kl (α) kk (β) ll = k (α) kk l (β) ll = χ (α) ˆ R χ (β) ˆ R (C.22) This rule can be naturally generalized for higher number of irreducible rep- resentations in the direct product (just multiply the characters of the irreducible representations). in a while we will have the product of three irreducible represen- tations. When is an integral bound to be zero? Everyone studying this book should know how to calculate the integral +1 −1 x dx = x 2 2 +1 −1 = 1 2 − 1 2 =0 Note, however, that we can tell what the value of the integral is without any calcu- lation, just by looking at the integrand. Indeed, the integrand is odd with respect to 4 Integrals important in spectroscopy 931 the transformation x →−x, i.e. the plot of the integral is an antisymmetric func- tion with respect to the reflection in the plane perpendicular to x at x = 0. The integration limits are symmetric with respect to that point. An integral means the area under the plot, therefore what we gain for x>0, we lose for x<0andthe integral will be exactly zero. The force of group theory relies, even with a complicated integrands, on being able to tell immediately whether the integral is equal to zero. This allows us to predict whether an optical transition is allowed or forbidden. We have to stress that these conclusions will be valid independent of the ap- proximations used to calculate the molecular wave functions. The reason is that they follow from the symmetry, which is identical for exact and approximate wave functions. The previous example can be generalized. Let us take the integral f α f β f γ dτ (C.23) where f α f β f γ transform according to the irreducible representations (α) (β) (γ) , of a symmetry group, and the integration is over the whole space. WHEN IS THE INTEGRAL EQUAL TO ZERO? If a representation (in general reducible), being the direct product of the irreducible representations (α) (β) (γ) does not contain the fully symmetric representation (its all characters are equal to 1), the integral is equal to zero. This is precisely our goal in this Appendix and this is why we have been working so hard with symmetry groups, operations, characters, etc. The essence of the theo- rem is very simple. The product f α f β f γ transforms according to the (in general reducible) representation, which is the direct product of the irreducible represen- tations (α) (β) (γ) . This means that according to eq. (C.17) the integrand f α f β f γ can be represented as a linear combination of the basis functions of all the irreducible representations: f α f β f γ = μ g μ ,whereg μ transforms accord- ing to the irreducible representation (μ) . Therefore, the integral (C.23) is the sum of the integrals f α f β f γ dτ = μ g μ dτ (C.24) Let us take one such integral: g μ dτ. Note that the integration is over the whole space (i.e. the integration limits are symmetric). If the integrand g μ were anti- symmetric with respect to one or more symmetry operations, the integral would automatically be zero (the same argument as for x dx). From this it follows that 932 C. GROUP THEORY IN SPECTROSCOPY all integrals in the sum would be zero except the single one which contains the integrand transforming according to the fully symmetric representation. 33 There are two (important to us) special cases of this theorem. Two special cases f α f β dτ = δ αβ A, i.e. in order for the integral not to vanish we have to have: (α) = (β) . The proof is very simple and relies on the fact that the characters of the fully symmetric irreducible representation are equal to 1. The number of times, a(A), the fully symmetric representation A, is present in the direct product (α) × (β) we calculate from the formula: a(A) = 1 g i χ (α×β) ˆ R i χ (A) ˆ R i ∗ = 1 g i χ (α×β) ˆ R i = 1 g i χ (α) ˆ R i χ (β) ˆ R i ∗ =δ αβ (C.25) This means that the fully symmetric representation is always present in (α) × (α) and therefore the integral does not vanish. 34 Letustaketheintegral f α f β f γ dτ (C.26) where f α f β f γ transform according to the irreducible representations α β γ. For the integral not to vanish the direct product (α) × (β) has to contain the representation (γ) . This means that to have integral (C.26) not vanish, the function f α f β decom- poses (eq. (C.17)) in such a way that there is a non-zero component belonging to (γ) . If this happens, according to the previous case, a component of the integrand will transform according to the fully symmetric representation, which will save the integral (C.26) from vanishing. Selection rules for electronic transitions (UV-VIS) The selection rules will be shown taking the example of pyrazine and its mono- and diprotonated ions (Fig. C.5). 33 Only for the fully symmetric representation are all the characters equal to 1, and therefore the corresponding function does not change under symmetry operations. 34 This is easy to understand. What transforms, according to (α) × (α) , is the product of two (in general different) functions, each belonging to (α) . It means that the function behaves in a very special way (typical for (α) ) under symmetry operations, e.g., changes sign under ˆ R 1 , while other operations leave it unchanged. If we have a product of two such functions, this means the product does not change at all under ˆ R 1 (and, of course, the other operations), i.e. transforms according to the fully symmetric operation. This is why the fully symmetric representation is always present in (α) × (α) . 4 Integrals important in spectroscopy 933 Fig. C.5. Pyrazine (a) and its mono- (b) and diprotonated (c) derivatives. The x axis is perpendicular to the ring plane, the y axis is in the ring plane perpendicular to the NN axis, and z is the NN axis. A glimpse at the chemical formulae is sufficient to tell that the monocation of pyrazine has the same symmetry as H 2 O, which corresponds to symmetry group C 2v (see Table C.4), while pyrazine and its diprotonated derivative have the sym- metry identical with that of naphthalene, i.e. D 2h . Let us focus first on the last case. Example 13. Pyrazine and its diprotonated derivative. Every book on group theory contains a table of characters of the symmetry group D 2h (Table C.6, x axis per- pendicular to the plane of the molecule, z goes through the nitrogen atoms). Table C.6. D 2h group table of characters D 2h ˆ E ˆ C 2 (z) ˆ C 2 (y) ˆ C 2 (x) ˆ ı ˆσ(xy) ˆσ(xz) ˆσ(yz) A g 11 1 1 1 1 1 1 x 2 y 2 z 2 B 1g 11 −1 −111−1 −1 R z xy B 2g 1 −11−11−11−1 R y xz B 3g 1 −1 −111−1 −11R x yz A u 11 1 1 −1 −1 −1 −1 B 1u 11 −1 −1 −1 −111z B 2u 1 −11−1 −11−11y B 3u 1 −1 −11−111−1 x 934 C. GROUP THEORY IN SPECTROSCOPY From Table C.6 we see 35 that what we call irreducible representations are sim- ply the distinct rhythms of pluses and minuses, which after squaring, give the fully symmetric behaviour. All the electronic states of pyrazine and its diprotonated derivative can be described by the irreducible representation labels: A g B 1g B 2g , B 3g A u B 1u B 2u B 3u . We may ask: what are the selection rules for state-to-state optical transitions? Are all transitions allowed, or are some of them forbidden? From the theory of the electromagnetic field (cf. Chapters 7 and 12) it follows that the probability of the transition between states k and l is proportional to |μ kl (x)| 2 or to |μ kl (y)| 2 or to |μ kl (z)| 2 , respectively, 36 depending on the electromagnetic wave polarization along axes x y or z axes, with: μ kl (x) = ψ ∗ k ˆμ x ψ l dτ μ kl (y) = ψ ∗ k ˆμ y ψ l dτ (C.27) μ kl (z) = ψ ∗ k ˆμ z ψ l dτ where ψ stands for the electronic states k,andl [eq. (6.8) on p. 225], ˆμ x ˆμ y ˆμ z are the operators of the molecular dipole moment components, 37 e.g., ˆμ z = i q i z i q i is the electric charge of the particle (electron or nucleus) having its z component equal to z i . Since we will decide, by using group theory, whether this integral 38 vanishes or not, the important thing is that μ x transforms in exactly the same way as the coordinate x.Theintegrandψ ∗ k ˆμ x ψ l transforms as the direct product of the three irreducible representations: that of ψ k ,thatof ˆμ x and that of ψ l . Excitations from the ground-state Suppose we • have a molecule in its ground-state ψ k (thus, belonging to the fully symmetric irreducible representation A g ), 35 Note that all the irreducible representations of the symmetry group of the molecules under consid- eration are one-dimensional, hence their energy levels are non-degenerate. 36 From the equality |μ kl (x)| 2 =|μ lk (x)| 2 and similarly for y and z it follows that the optical excita- tion and the corresponding emission have the same probability. 37 This may look alarming, because the operator depends on the choice of the coordinate system (cf. Appendix X). Do not worry, everything is all right. Even if the dipole moment depends on such a choice, any two choices give dipole moments differing by a constant vector. This vector being a constant can be shifted outside the integral and the integral itself will become zero, because ψ k and ψ l are orthogonal. Thus, to our delight, light absorption does not depend on the choice of the coordinate system. This is fine. 38 The integration goes over all the electronic coordinates. 4 Integrals important in spectroscopy 935 • immobilize the molecule in space (say, in a crystal), • introduce the coordinate system in the way described above, • irradiate the molecule with light polarized along the x axis and ask which states will the molecule be excited to. The direct product of A g and polarization x the irreducible representation to which x belongs, decomposes into some irre- ducible representations. For the optical transition to be allowed, we have to find among them the irreducible representation to which ψ l belongs (just recall that f α f β dτ = δ αβ A). Only then will the integrand contain something that has the chance to transform according to the fully symmetric representation. The x coor- dinate belongs to the representation B 3u (see Table C.6, last column). Therefore, let us see what represents the direct product A g ×B 3u .Wehaveeq.(C.12)forthe integer a(α) that is the number of irreducible representations α in a given reducible representation. Let us calculate this number for the representation (in general re- ducible) being the direct product, and all the irreducible representations α.Inthis particular case the direct product is 39 A g ×B 3u .Wehave a(A g ) = 1 8 [1 ×1 +1 ×(−1) +1 ×(−1) +1 ×1 +1 ×(−1) +1 ×1 +1 ×1 +1 ×(−1)]=0 a(B 1g ) = 1 8 [1 ×1 +1 ×(−1) +(−1) ×(−1) +(−1) ×1 +1 ×(−1) +1 ×1 +(−1) ×1 +(−1) ×(−1)] = 0 etc., all zeros, and finally a(B 3u ) = 1 8 [1 ×1 +(−1) ×(−1) +(−1) ×(−1) +1 ×1 +(−1) ×(−1) +1 ×1 +1 ×1 +(−1) ×(−1)]=1 exactly as we have expected. Thus, we can write 40 A g ×B 3u =B 3u 39 The characters of A g × B 3u are as follows (in order of the symmetry operations in the table of characters): 1 −1 −11−111−1 i.e. they are identical to those of the (it turns out irreducible) representation B 3u . Such a product is really easy to form. In the table of characters one finger goes horizontally over the characters of A g (they are all equal to 1), while the second finger moves similarly over the characters of B 3u and we multiply what the first finger shows by what the second finger shows. The result is the character of the direct product A g ×B 3u , which in this case turns out to be exactly the character of B 3u . This is why we may expect that a(α) will all be zero except a(B 3u ) =1. 40 We may say that the fully symmetric representation plays the role of unity in the multiplication of irreducible representations. . (γ) ki are the matrix elements of the irreducible representation γ Z ijkl = (α) ki (β) lj . Of course, the dimension of this representation is the product of the dimensions of the representations. function that is a linear combination of the basis functions of a reducible representation can be decomposed into a linear combination of the basis functions of those irreducible representations. form the basis of an irreducible representation of the symmetry group of the molecule, or in other words, transform according to this irreducible representation – the dimension of the representation
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