Symmetry Labeling of Molecular Energies

Capter 7. Symmetry Labeling of Molecular Energies Notes: Most of te material presented in tis capter is taken from Bunker and Jensen 1998, Cap. 6, and Bunker and Jensen 2005, Cap. 7. 7.1 Hamiltonian Symmetry Operations We ave seen in section 6.1.4 of te previous capter tat any elements of te CNPI group, and terefore any elements of te MS group, associated wit a given molecule commute wit te corresponding molecular Hamiltonian. Tat is, given a symmetry operator we ave!ĥ 0, = 0. 7.1 We also saw tat tis implies tat if a wave function! of te Hamiltonian is transformed by suc tat! =!, ten tis transformed function is also a wave function of te Hamiltonian wit te same energy level as te original function, since Ĥ 0! n = Ĥ 0! n = Ĥ 0! n = E n 0! n = E n 0! n, 7.2 were E 0 n is te energy associated wit te wave function! n. For a non-degenerate energy level, tis in turn implies tat! n = c! n, 7.3 wit c some constant. As we will now see, te determination of te different values tat tis constant can take is central to te use of symmetry labels, associated wit eac irreducible representation of te MS group, for identifying te energy levels of a molecule. 7.1.1 Non-degenerate Energy Levels Consider again te case of a MS group operator acting on a wave function! n of nondegenerate energy level E n 0. However, because of te group axioms tat states tat bot q, wit q and integer, and te identity E must be part of te group, it follows tat for some integer m we must ave We terefore find from equation 7.3 tat m = E. 7.4 m! n = c m! n = E! n =! n, 7.5 119

or alternatively For example, for te permutation 12 we find tat c = ±1. Example c = m 1. 7.6 Consider te following tree fictitious wave functions for te water molecule te protons are labeled 1 and 2! 1 = sin X 1 X 2! 2 = cos X 1 X 2! 3 = sin X 1 + X 2. 7.7 Determine te constants associated wit eac operator of te MS group for tis molecule, and use te corresponding caracter table to determine te irreducible representation generated by te wave functions. Solution Te MS group for tis molecule is C 2v M Applying te different operators to te wave functions we ave and its caracter table is given in Table 7-1. =! 1 =! 1 =! 1! 1 E = sin X 1 X 2 12! 1 = sin X 2 X 1! 1 E = sin X 2 X 1 12! 1 = sin X 1 X 2 =! 1, 7.8 and similarly! E 12 2 =! E 2 =! 12 2 =! 2 =! 2! E 12 3 =! E 3 =! 12 3 =! 3 =! 3. Table 7-1 Te caracter table of te C 2v M group. C 2v M : E 12 E! 12! A 1 : 1 1 1 1 A 2 : 1 1!1!1 B 1 : 1!1!1 1 B 2 : 1!1 1!1 7.9 120

It is terefore apparent tat te caracters of te different wave functions wen acted upon wit te MS group operators are suc tat if we collect te different constants c i, wit i = 1,2,3 for te wave functions, we ave c i E 12 c i E c! 12 i c! i! 1 : 1!1!1 1 = B 1! 2 : 1 1 1 1 = A 1! 3 : 1 1!1!1 = A 2 And, as indicated in te last column,! 1,! 2, and! 3 ave te symmetry of B 1, A 1, and A 2, respectively. Because te energy levels are different for te tree wave functions tey are non-degenerate, ten we find tat we can label te energy levels using te different irreducible representations of te MS group. 7.1.2 Degenerate Energy Levels Wen an energy level E 0 n is l-fold degenerate, tere are l wave functions! n1,,! nl tat transform into linear combinations of temselves wen acted upon by an operator of te MS group. Tis is expressed matematically as follows l! ni = D[ ] ij! nj, 7.10 j =1 were D[ ] is te representation matrix associated to. Now imagine tat we are forming anoter set of l-fold degenerate functions! nk using an ortogonal matrix A wit l! nk = A ki ni. 7.11 i=1 Evidently tese new functions are also wave functions of te Hamiltonian wit te same 0 energy E n as te original wave functions! ni. We would like to find out wat representation of te MS group te wave functions! nk generate. If we denote te new representation matrix associated to wit D[ ], ten But using equations 7.11 and 7.10 we can also write l! nk = D[ ] kr! nr. 7.12 r =1 121

l! nk = A ki ni i=1 l l = A ki D[ ] ij nj, i=1 j =1 7.13 and we substitute for! nj using te inverse of equation 7.11 to get l! nk = A ki D i=1 l j =1 [ ] ij A jr l r =1 1! nr 1 = A ki D[ ] ij A jr! nr r l i, j kr! nr = AD[ ]A 1. r =1 7.14 [ ] is [ ] via a similarity transformation. Tat is, Comparing tis result wit equation 7.12 reveals tat te new representation D related to te original representation D D[ ] = AD[ ]A!1. 7.15 Since we already know tat te caracter of a matrix is invariant under a similarity transformation see equation 6.50 of te previous capter, ten we find te important result tat te caracter of a representation generated by a set of degenerated wave functions of te Hamiltonian is unique and can be used to reduce it to its irreducible components see equations 6.59 and 6.60 of Capter 6. Te corresponding degenerated energy level can tus be labeled wit te irreducible representations of te MS group. Tis result is an extension of wat was previously found for non-degenerate energy levels. 7.2 Projection Operators Let us again consider a set of wave functions! js tat generate te irreducible representations! j of dimension i.e., degeneracy l j, wit s = 1,,l j. Any oter set of similar size of functions! n generating a reducible representation! can expressed as a linear combination of te! js because tey form a basis wit l j k =1! n = A n, jk jk. 7.16 j Note: Te notation of equation 7.16 is potentially confusing, but it is understood tat A n, jk is te element of a matrix A belonging to row n and column jk. Tat is, te indices j and k taken togeter form a single index. Te implication is tat te column 122

vector representation of! jk is suc tat its first l 1 elements belong to j = 1 and k = 1,,l 1, te next l 2 elements belong to j = 2 and k = 1,,l 2, and so on. Equation 7.16 can also be inverted to give Now consider te following operator! jk = A 1 jk,n n. 7.17 n were D! i! P i mm = l i D! i [ ] mm 7.18 [ ] is te matrix associated to te operator in te! i irreducible representation of te corresponding symmetry group of order. We now apply tis operator to a function of te! n set! P i mm n = l i = l i l j k =1 D! i D! i [ ] mm n [ ] mm = A n, jk l i j l j k =1 = A n, jk l i j l j k =1 l j s=1 & D! i D! i j l j k =1 [ ] mm & [ ] mm A n, jk % jk + * % jk l j D! j s=1 [ ] ks % js + * & l = A n, jk % i js D! i [ ] mm D! j [ ] ks * +, j 7.19 were equations 7.16 and 7.10 suitably rewritten were used. We can now make use of te GOT i.e., equation 6.54 of Capter 6 to find tat P mm % l j % k =1 l j % s=1! i n = A n, jk js ij mk ms j = A n,im im. 7.20! We terefore find tat te projection operator P i mm wen applied to! n produces te part of! n tat belongs to te m t row of te! i irreducible representation. In oter [ ] of a group we can use te words, given te matrix irreducible representations D! i projection operator to break down an arbitrary set of functions! n into te irreducible 123

representations-generating wave functions! im tat compose it. For example, if te functions! n do not contain te! i irreducible representation, ten equation 7.20 will yield! P i mm n = 0. 7.21 It is important to empasize te fact we need to know te matrices D! i [ ] in order to use tis projection operator. On te oter and, is possible to define a simpler and less powerful projection operator from equation 7.18 wit For example, we can calculate from equation 7.20 tat P! i! = P i mm = l i! i [ ] 7.22 m P! i = A n n,im im. 7.23 m So, wen applied to! n tis projection operator will yield a function tat belongs to! i, but not belonging to any particular row m. However, it will still be able to tell us if! n does not contain any functions belonging to! i, since in tis case P! i n = 0, 7.24 from equation 7.21. It is important to note tat for a one-dimensional irreducible representation te operators of equations 7.18 and 7.22 are one and te same. Example Let us return to our previous example of te water molecule, but tis time wit tree new wave functions defined as + sin X 1 + X 2 + cos X 1 X 2 sin X 1 X 2,! a = sin X 1 X 2 = sin X 1 X 2! c = sin X 1 + X 2 7.25 wic we assume to be degenerate i.e., we now specify tat! 1,! 2, and! 3 are also degenerate and spanning a reducible representation!. We seek to determine te combinations of! a,, and! c tat belong to te different irreducible representations of te MS group of te water molecule i.e., C 2v M. 124

Solution Using te caracter table for C 2v M i.e., Table 7-1 we can calculate te 3! 3 matrices D! [ ] corresponding to te transformation brougt by te operators of te group on! a,, and! c. For example, we can verify from equations 7.25 tat E 12 E *! a! c! a! c! a! c % = &! a! c % 1 0 0% = 0 1 0 & 0 0 1&!! c [ ] D E % = +! c! a &! a! a! c %. & % 0 0 1% = 1 1 1 & 1 0 0&! D 12 %& %! a % 1 0 0 % = +! c! a = 1 1 1 &! c & 0 0 1&! D E * % &! a! c % &! a! c %, & 7.26 and D! 12 % & = D! 12 %& D! E % & 0 0 1% 1 0 0 % 0 0 1% = 1 1 1 1 1 1 = 0 1 0. 1 0 0& 0 0 1& 1 0 0 & 7.27 Te caracters of tese matrices are easily determined to be [ ] = 3,! 12! E & % = 1,! E& % = 1,! 12 Using equation 6.60 of Capter 6 and Table 7-1, we find tat a A1 = 1 4! [ ]! A 1 [ ] % = 1. 7.28 7.29 = 1 = 1 4 3 + 1 % 1 + 1 125

a A2 = 1 4! [ ]! A 2 [ ] = 1 4 3 + 1+ 1 % 1 = 1 a B1 = 1 4! [ ]! B 1 [ ] = 1 4 3 % 1+ 1 + 1 = 1 7.30 a B2 = 1 4! [ ]! B 2 [ ] = 1 4 3 % 1% 1 % 1 = 0. We can terefore write! = A 1 A 2 B 1, 7.31 te B 2 representation is not generated by!. Te functions belonging to te tree realized irreducible representation are determined troug equation 7.22 and te caracter table of te MS group. For example, using! a we ave P A 1! = % 1 a & 4 A 1 [ ] = 1 4! +! +! +! a c a c = 0 *! = 1 a { 4 E + 12 + E + 12 }! a P A 2! = % 1 a & 4 P B 1! = % 1 a & 4 A 2 [ ] *! = 1 a { 4 E + 12 + E + 12 }! a = 1 4! +! +! +! a c a c = 1 2! +! a c B 1 [ ] *! = 1 a { 4 E + 12 + E + 12 }! a = 1 4! +! +! +! a c a c = 1 2! +! a c 7.32 P B 2! = % 1 a & 4 B 2 [ ] = 1 4! +! +! +! a c a c = 0. *! = 1 a { 4 E + 12 + E + 12 }! a 126

Applying P! i to! c would yields te same results. Finally, projecting onto eac irreducible representation gives P A 1! = % 1 b & 4 P A 2! = % 1 b & 4 A 1 [ ] *! = 1 b { 4 E + 12 + E + 12 } + +! c +! a + = 1 4,-! +! +! + b c a. / = + 1 2! +! c a A 2 [ ] *! = 1 b { 4 E + 12 + E + 12 } P B 1! = % 1 b & 4 + +! c +! a + = 1 4,-! +! +! + b c a. / = 0 B 1 [ ] *! = 1 b { 4 E + 12 + E + 12 } + +! c +! a + = 1 4,-! +! +! + b c a. / = 1 2! +! a c 7.33 P B 2! = % 1 b & 4 B 2 [ ] *! = 1 b { 4 E + 12 + E + 12 } + +! c +! a + = 1 4,-! +! +! + b c a. / = 0. Summarize tese results, we find tree functions! A 1 = b + 1 2 c a = cos X 1 X 2! A 2 = 1 2 + a c = sin X 1 + X 2 7.34! B 1 = 1 2 a c = sin X 1 X 2, wic respectively transform as te A 1, A 2, and B 1 representations. A careful look at tese functions reveals tem to be equaled to our previous! 2,! 3, and! 1, respectively, wic we also previously determined to be transforming as te A 1, A 2, and B 1 representations. Tis is an example of te power of te projection operator for identifying underlying symmetries in oterwise arbitrary functions. 7.2.1 Te Symmetry of a Product of Functions As we saw earlier wen dealing wit te molecular Hamiltonian and its simplifications e.g., te Born-Oppeneimer approximation, te complete wave functions corresponding to te rovibronic states of a molecule can be approximated by te product of electronic, 127

vibration, and rotational wave functions. It is terefore important to determine te symmetry and caracter of a product of functions. Let us consider two representations! m and! n of r-fold and s-fold degeneracy, respectively. We also assume tem to be generated by corresponding sets of wave functions! mi and! nj i.e., i = 1,,r and j = 1,,s of energy levels E m and E n, respectively. We already know from equation 7.10 tat under an element of te MS group functions from tese sets will transform as follows r! mi = D m k =1 s [ ] ik! mk! nj = D [ n ] jl! nl. l =1 7.35 If we denote te representation generated by all possible r! s products of pairs of functions! mi,! nj by! mn but if we furter define ten we can write r s! mi! nj % = [ ] D&m ik D [ &n ] jl! mk! nl, 7.36 k =1 l =1 We identify and define te matrix D!mn! ij = mi nj, 7.37 r s! ij = D [ mn ] ij,kl! kl. 7.38 k =1 l =1 [ ] by te direct product D!mn [ ] = D [!m ] D [!n ], 7.39 or D!mn [ ] ij,kl = D [!m ] ik D [!n ] jl. 7.40 For example, we represent te direct product of two 2! 2 matrices A and B wit 128

a 11 b 11 a 11 b 12 a 12 b 11 a 12 b 12 % A! B = a 11 a 12 % a 21 a 22 &! b 11 b 12 % b 21 b 22 & = a 11 b 21 a 11 b 22 a 12 b 21 a 12 b 22. 7.41 a 21 b 11 a 21 b 12 a 22 b 11 a 22 b 12 a 21 b 21 a 21 b 22 a 22 b 21 a 22 b 22 & We see tat te first and tird indices of D!mn [ ] ij,kl specify te quadrant of te resulting matrix, wile te second and fort indices specify wic element of te quadrant is selected. From tis definition we can easily proceed to calculate te caracter of te direct-product matrix wit! mn r s [ ] = D [ mn ] ij,ij = D m k =1 l =1 =! [ m ]! [ n ]. r s i=1 j =1 [ ] ii D [ n ] jj 7.42 Te caracter of a direct product matrix is tus simply te product of te caracters of te two matrices involved. We also write symbolically Example Consider te two pairs of degenerate functions! mn =! m! n. 7.43! 1 = X 1 X 2! 2 = 1 3 2X 3 X 1 X 2, 7.44 and! a = Y 1 Y 2 = 1 3 2Y Y Y 3 1 2. 7.45 and terefore! a, generates te E representation of te M group see Table 7-2. a Sow tat! 1,! 2 C 3v b Determine te representation generated by te products! 1,! 2 decompose it into a sum of te irreducible representations of C 3v M. a, b, and 129

Table 7-2 Te caracter table of te C 3v M group. Solution. C 3v M : E 123 132 23! 13! 12! A 1 : 1 1 1 A 2 : 1 1!1 E : 2!1 0 a Applying one element from eac of te tree classes of operators from te C 3v M group to te functions! 1 and! 2 we ave E! 1 =! 1 E! 2 =! 2 123! 1 = X 3 X 1 = 1 2! 3! 1 2 = 1 2 123! 2 = 1 3 2X 2 X 3 X 1 23! 1 = X 1 + X 3 = 1 2! 1 3! 2 3! 1 +! 2 7.46 From tis, we easily find tat 23! 2 = 1 3 2X X X 2 1 3 = 1 2 [ ] = 2,! 123! E %& = 1,! 23 3! 1 +! 2. % & = 0, 7.47 wic makes it clear, troug a comparison wit Table 7-2, tat! = E. Tat is, te set! 1,! 2, and terefore! a, also, generates te E representation of te C 3v M group. b We know from equations 7.42, 7.43, and 7.47 tat! E [ ] = 4,! 123 Using equation 6.60 of te previous capter % %& = 1,! & 23 = 0. 7.48 a i = 1 &! [ ]! [ i ] %, 7.49 6 130

we find + 3 1! 0 a A1 = 1 1! 4 + 2 1!1 6 % = 1 + 3 &1! 0 a A2 = 1 1! 4 + 2 1!1 6 % = 1 7.50 + 3 0! 0 a E = 1 6 2! 4 + 2 &1!1 % = 1, or alternatively! = A 1 A 2 E. 7.51 If we now apply te projection operators see equation 7.22 for te A 1 and A 2 irreducible representations to some of te product functions! i we could sow troug some simple matematics tat P A 1! 1 a! 1 a +! 2 b P A 2! 1 b! 1 b! 2 a. 7.52 In a similar manner, if we use equation 7.18 for te projection operator for, and equations 6.44 for te matrices of, te E irreducible representation of C 3v M it can be sown tat P E 11! 1 a! 1 a! 2 b P E 22! 1 b! 1 b +! 2 a. 7.53,! 1 b! 2 a, and te pair of functions Tat is, te functions! 1 a +! 2 b %! 1 a! 2 b,! 1 b +! 2 a & respectively generate te A 1, A 2, and E irreducible representations. In oter words, te symmetric product functions! 1 a,! 2 b, and! 1 b +! 2 a generate te representation A 1! E, wile te antisymmetric product generate te representation A 2. functions! 1 b! 2 a It is usually written tat te symmetric product is and tat te antisymmetric product is [ E] 2 = [ E! E] = A 1 E, 7.54 { E} 2 = { E! E} = A 2. 7.55 131

It can be sown 1 tat te caracter of te operators in te symmetric and antisymmetric product representations can be calculated wit! [ EE] [ ] = 1 2! { EE} [ ] = 1 2! E [ ] 2 +! E 2 %% & &! E [ ] 2! E 2 7.56 %% & &, as can be verified for te preceding example. In general, te caracter of te operators in te symmetric nt power of te degenerate irreducible representation E is given by! [ E]n [ ] = 1 2! E [ ]! E [ ]n1 [ ] +! E n %% & &. 7.57 An important question wen considering a representation! resulting from te direct products of functions generating two irreducible representations! n and! m is weter it contains te totally symmetric representation! s! n is te irreducible representation wose matrix representation D! n D! n [ ]. We know from equation 7.42 tat and te from equation 6.60 tat [ ] is te complex conjugate of tat of! n, i.e.,! [ ] =! n [ ]! m [ ], 7.58 a! s = 1! [ ]. 7.59 We now insert equation 7.58 in equation 7.59 to get a! s = 1! n [ ]! m [ ], 7.60 wic implies from te little ortogonality teorem i.e., equation 6.56 tat a! s = 1, if! =! m n % 0, if! m! n, 7.61 since if! m is a realized irreducible representation, ten so is! n. 1 M. Hamermes 1964, Group Teory and its Applications to Pysical Problems, eading: Addison-Wesley, pp. 128-134. 132

7.3 Te Vanising Integral ule Let us suppose tat we ave a molecular Hamiltonian Ĥ 0 for wic we ave previously determined te energy levels E 0 n and corresponding wave functions! 0 n. We assume tat te wave function! n 0 generate te representation! n. We would like to know wic of te molecular states and energy levels will be coupled if a perturbation H ˆ! is added to te system. In oter words, will te perturbation allow te molecule to go from a given state to anoter? Tis question will be answered by calculating te matrix elements of te total Hamiltonian Ĥ = Ĥ 0 + H! wit 0 H mn = %! m Ĥ 0 + H ˆ! 0 n d = & mn E 0 n + H mn, 7.62 were 0 H mn! = % ˆ m H! 0 n d, 7.63 wit d! te volume element of te relevant space. Tat is, te matrix element H mn! as to be non-zero for a transition to be possible between two different states m and n. It turns out tat we can easily find out weter te integral in equation 7.63 vanises or not by making use of te fact tat molecular energies can be labeled wit te irreducible representations of te MS group see Section 7.1. To sow ow tis is done, let us define a function f suc tat 0 f S =! ˆ m H! 0 n, 7.64 wit S te coordinates of some point in te system s space. If we act on tis function wit some element i of te group G of order, ten we can write see equation 6.9 i f S = f S! = f i S But since te integration in equation 7.63 is performed over all space and terefore. 7.65 f Sd! = f S d! = f i Sd! = % i f S & d!, 7.66 f Sd! = & i f S % i d!. 7.67 133

We can ten rewrite equation 7.63 as 0 H mn! = % ˆ m H! 0 n d = 1 0 i ˆ * & H m! 0 % n, i + d. 7.68 If te perturbation H ˆ! generates te representation!, ten we know from te previous 0 section tat! ˆ m H! 0 n will generate te following representation! mn = m! n. 7.69 0 Note tat! m generates! m, wic does not necessarily ave caracters tat are complex conjugate of tose of! m. Let us now apply te projection operator for te totally symmetric representation P! s = 1 i 7.70 on te integrand of equation 7.63 to find tat i & P! s 0 ˆ m H 0 n d% = 1 0 i ˆ + H m 0 & * n - i, d% 13 = 0, if! s. = H mn 2 43 / 0, if! s 0! mn! mn 7.71 see Section 7.2. We terefore find tat and te perturbation H ˆ! will not couple two states wen 0 H mn! = % ˆ m H! 0 n d = 0 7.72! s! m!%! n. 7.73 Tis is te so-called vanising integral rule. In particular, te condition spelled out in equations 7.72 and 7.73 reduces to! m! n 7.74 wen H ˆ! is totally symmetric i.e., transforms as! s, see below in te group G. 0 If te energy levels E m are r-fold degenerate wit wave functions! mi for i = 1,,r and wen H ˆ! is totally symmetric, ten te previous result can be furter complemented wit a judicious use of te GOT; see te Second Problem List wit te following rule %! 0 H ˆ! 0 mi d = 0 7.75 nj 134

unless! m =! n and i = j. Incidentally, we note tat te fact tat from part of equation 7.62 0 Ĥ mn 0 =! Ĥ 0 m! 0 0 n d = % mn E n 7.76 implies tat te corresponding element of te unperturbed Hamiltonian matrix will cancel unless! m =! n. Tis not a surprising result since we built te basis! m 0 { } from a set of ortogonal wave functions of tis Hamiltonian. However, tis also means tat te Hamiltonian Ĥ 0 transforms as te totally symmetric representation! s of te MS group. We could ave easily predicted tis from te fact tat Ĥ 0 commutes wit te operators of te MS group see equation 7.1, and is terefore invariant under teir applications. Example Consider te two wave functions = sin X 1 + X 2 = sin X 1 X 2! A 2! B 1 7.77 introduced in an earlier example concerning te C 2v M integral of tese functions over all space will cancel out. Tat is, group. It is easy to see tat te sin X 1 + X 2 dx 1 dx 2 sin X 1! X 2 dx 1 dx 2 = 0, 7.78 =!!! since bot integrands are odd. We can also explain tis result wit te vanising integral rule by saying tat tese functions are not totally symmetric in te C 2v M group, were! s = A 1, since tey transform as A 2 and B 1, respectively. Likewise, te product of te two functions generates te A 2! B 1 = B 2 irreducible representation see Table 7-1 and its integral will also cancel out for te same reason. Tis can also be verified wit te following! I = sin X 1 + X 2 sin X 1! X 2 dx 1 dx 2!!!! cos 2X 1 = 1 cos 2X 2 2 % & dx 1 dx 2 = 0.! 7.79 On te oter and, te integral of te squares of bot functions will not cancel out because A 2! A 2 = B 1! B 1 = A 1 = s, 7.80 135

as can also be verified wit I 1 = sin 2 X 1 + X 2 dx 1 dx 2!! = 1 1! cos 2 X 1 + X 2 2 { % & }dx 1 dx! 2 0! I 2 = sin 2 X 1! X 2 dx 1 dx 2!! = 1 1! cos 2 X 1! X 2 2 { % & }dx 1 dx! 2 0.! 7.81 It can be seen tat te vanising integral rule is closely related to te more common concept of te evenness or oddness of te integrand. In general, te vanising integral rule cannot tell us wat is te value of an integral, but it can tell us wen it cancels out. And tis is of fundamental importance for determining te selection rules for transitions between different molecular states. 136