Slater-Condon Rules. Antisymmetrization Operator APPENDIX M

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1 APPENDIX M Slater-Condon Rules The Slater determnants represent somethng lke the daly bread of quantum chemsts. Our goal s to learn how to use the Slater determnants when they are nvolved n calculaton of the mean values or the matrx elements of some mportant operators. We wll need that n the Hartree-Fock method, as well as n other mportant methods of quantum chemstry. Most mportant are only the fnal results of the dervatons presented n ths appendx (p. e119). Antsymmetrzaton Operator The antsymmetrzaton operator s defned as  = 1 ( 1) p ˆP, N! P (M.1) where ˆP represents a permutaton operator of N objects (electrons, n our case), whle ( 1) p stands for the party of the permutaton P, even ( odd ), f a gven permutaton P can be created from an even (odd) number p of the transpostons (.e., exchanges) of two elements. The operator  has some nce features. The most mportant one s that when appled to any functon, t produces ether a functon that s antsymmetrc wth respect to the permutatons of N elements, or zero. 1 Ths means that  represents a sort of magc wand: whatever t touches becomes antsymmetrc or dsappears! The antsymmetrzer s also dempotent;.e., t does not change any functon that s already antsymmetrc, whch means  = Â. Let us check that  s ndeed dempotent. Frst, we obtan  = (N!) 1 P ( 1) p ˆP(N!) 1 P ( 1) p ˆP = (N!) PP ( 1) p+p ˆP ˆP. (M.) Of course, ˆP ˆP represents a permutaton operator, whch s then multpled by ts own party ( 1) p+p, and there s a sum over such permutatons at a gven fxed ˆP. Independent of what 1 In the near future, these elements wll be dentfed wth the electronc coordnates (one element wll be represented by the space and spn coordnates of a sngle electron: x, y, z,σ). The permutatons form the permutaton group. Note ths excerpt from Sold State and Molecular Theory, Wley, London (1975) by John Slater on the permutaton group: (...) It was at ths pont that Wgner, Hund, Hetler, and Weyl entered the pcture, wth ther Ideas of Quantum Chemstry, Second Edton Elsever B.V. All rghts reserved. e109

2 e110 Appendx M ˆP s, we obtan the same result 3 N! tmes, and therefore  = (N!) N! P ( 1)p ˆP = Â. Ths s what we wanted to show. The operator  s Hermtan. Snce ˆP represents a (permutatonal) symmetry operator, t conserves the scalar product. Ths means that for the two vectors ψ 1 and ψ of the Hlbert space, we obtan 4 ψ 1 (1,,...,N) Âψ (1,,...,N) = (N!) 1 ( 1) p ˆP 1 ψ 1 (1,,...,N) ψ (1,,...,N). P The summaton over ˆP can be replaced by the summaton over ˆP 1 : (N!) 1 ( 1) p ˆP 1 ψ 1 (1,,...,N) ψ (1,,...,N). P 1 Snce the party p of the permutaton ˆP 1 s the same as that of ˆP, hence (N!) 1 P 1 ( 1)p ˆP 1 = Â, whatshowsthatâ s Hermtan: ψ 1 Âψ = Âψ1 ψ,or 5  = Â. (M.3) Gruppenpest : the pest of the group theory, as certan dsgruntled ndvduals who had never studed group theory n school descrbed t. (...) The authors of the Gruppenpest wrote papers, whch were ncomprehensble to those lke me who had not studed group theory (...). The practcal consequences appeared to be neglgble, but everyone felt that to be n the manstream of quantum mechancs, we had to learn about t. (...) It was a frustratng experence, worthy of the name of a pest. 3 Of course, ˆP ˆP = ˆP has the party ( 1) p+p, because ths s how such a permutaton party s to be calculated: frst, we make p transpostons n order to get ˆP, and next, makng p transpostons, we obtan the permutaton ˆP ˆP. Note, that when keepng ˆP fxed and takng ˆP from all possble permutatons, we are runnng wth ˆP ˆP over all possble permutatons as well. Ths s because the complete set of permutatons s obtaned ndependently of what the startng permutaton looks lke (.e., ndependent of ˆP ). 4 The conservaton of the scalar product ψ 1 ψ = ˆPψ 1 ˆPψ means that the lengths of the vectors ψ 1 and ˆPψ 1 are the same (smlarly wth ψ ), and that the angle between the vectors s also conserved. If ˆP s actng on ψ alone and ψ 1 does not change, the angle resultng from the scalar product ψ 1 ˆPψ s of course dfferent because only one of the vectors (ψ ) has been transformed (whch means a rotaton of a unt vector n the Hlbert space). The same angle would be obtaned f ts partner ψ 1 were transformed n the opposte drecton;.e., when the operaton ˆP 1 ψ 1 had been performed. Hence, from the equalty of the angles, we have ψ 1 ˆPψ = ˆP 1 ψ 1 ψ. 5  stands for the adjont operator wth respect to Â;.e., for arbtrary functons belongng to ts doman, we have ψ  1 Âψ = ψ 1 ψ. There s a subtle dfference (gnored n ths book) among the self-adjont ( = Â) and Hermtan operators n mathematcal physcs (they dffer by defnton of ther domans).

3 Slater-Condon Rules e111 Slater-Condon Rules The Slater-Condon rules serve for expressng the matrx elements nvolvng the Slater determnants (they are many-electron wave functons): = 1 φ 1 (1) φ 1 () φ 1 (N) φ (1) φ () φ (N) N!. (M.4) φ N (1) φ N () φ N (N) The normalzed Slater determnant has the form: = N!Â(φ 1 φ φ N ),where φ 1 φ φ N represents a product φ 1 (1)φ () φ N (N), and therefore, the normalzaton constant before the determnant tself det[φ 1 (1)φ () φ N (N)] s equal to (N!) 1/. Quantum chemsts love the Slater determnants because they are bult of the one-electron brcks φ called spnorbtals (we assume them to be orthonormal) and because any Slater determnant s automatcally antsymmetrc wth respect to exchange of the coordnates of any 1 two electrons (shown as arguments of φ ), the factor ensures the normalzaton. At the N! same tme, any Slater determnant automatcally satsfes the Paul excluson prncple, because any attempt to use the same spnorbtals results n two rows beng equal, and as a consequence, n havng = 0 everywhere. 6 Usng the Slater determnants gve quantum chemsts some comfort snce all the ntegrals that appear when calculatng the matrx elements of the Hamltonan are relatvely smple. The most complcated ones contan the coordnates of two electrons. What Knd of Operators we wll be Dealng wth? 1. The sum of one-electron operators ˆF = ĥ(). The sum of two-electron operators Ĝ = < j ĝ(, j) In both cases, the summaton goes over all the electrons. Note that ĥ has the dentcal form ndependent of the partcular electron; and the same pertans to ĝ. The future meanng of the ˆF and Ĝ operators s obvous; the frst pertans to the nonnteractng electrons (electronc knetc energy wth ĥ() = 1 or the nteracton of the electrons wth the nucle), the second operator deals wth the electronc repulson, wth ĝ(, j) = 1 r j. 6 Ths s a knd of catastrophe n theory: because our system s somewhere and can be found there wth a certan nonzero probablty.

4 e11 Appendx M What the Slater-Condon Rules are all About? The Slater-Condon rules show how to express the matrx elements of many-electron operators ˆF and Ĝ wth the Slater determnants by the matrx elements of the operators ĥ and ĝ computed wth the orthonormal spnorbtals φ. The operators ˆF and Ĝ are nvarant wth respect to any permutaton of the electrons (Chapter ). In other words, the formulas for ˆF and Ĝ do not change before and after any relabelng of the electrons. Ths means that any permutaton operator commutes wth ˆF and Ĝ. Snce s a lnear combnaton of such commutng operators, then  ˆF = ˆF  and ÂĜ = Ĝ Â. A Smple Trck Used n the Proofs All the proofs gven here are based on the same smple trck. Frst, the ntegral under consderaton s transformed nto a sum of the terms φ 1 (1)φ () φ N (N)  ˆX φ 1 (1)φ () φ N (N), where ˆX = ĥ() or ĝ(, j). Then, we recall that  s a lnear combnaton of the permutaton operators, and that n the ntegral φ 1 (1)φ () φ N (N) ˆX φ n1 (1)φ n () φ nn (N), only a few terms wll survve. In case ˆX = ĥ(), we obtan a product of one-electron ntegrals: φ 1 (1)φ () φ N (N) ˆX φ n1 (1)φ n () φ nn (N) = φ 1 (1) φ n1 (1) φ () φ n () φ () ĥ() φ n () φ N (N) φ nn (N). Snce the spnorbtals are orthonormal, then only one term wll survve, the one that has (n 1, n,...,n 1, n +1,...n N ) = (1,,..., 1, + 1,...N). All the overlap ntegrals that appear there are equal to 1. Only one of the one-electron ntegrals wll gve somethng else: φ () ĥ() φ n (), but also n ths ntegral, we have to have n = because of the abovementoned overlap ntegrals that force the matchng of the ndces. If ˆX =ĝ(, j), we make the same transformatons, but the rule for survval of the ntegrals pertans to the two-electron ntegral that nvolve the coordnates of the electrons and j (not one-electron as before). Note that ths tme we wll have some pars of the ntegrals that are gong to survve, because the exchange of the ndces j j also makes an ntegral that survves.

5 Slater-Condon Rules e113 I Slater-Condon Rule If ψ represents a normalzed Slater determnant, then N F = ψ ˆF ψ = ĥ, =1 G = ψ Ĝ ψ = 1 ( j j j j ),, j (M.5) (M.6) where ĥ r σ 1 φ (1)ĥ(1)φ r (1)dV 1 (M.7) j kl φ (1)φ j ()g(1, )φ k(1)φ l ()dv 1 dv, σ 1 σ where the summaton pertans to two spn coordnates (for electrons 1 and ). Proof : Operator ˆF F = ψ ˆF ψ =N! Â(φ 1 φ φ N ) ˆF Â(φ 1 φ φ N ) Usng  ˆF = ˆF Â,  =  and  = Â, one gets (M.8) F = N! φ 1 φ φ N Â[(ĥ(1)φ 1 φ φ N ) + (φ 1 φ ĥ(n)φ N )] = N! N! φ 1φ φ N ˆ1[(ĥ(1)φ 1 φ φ N ) + (φ 1 φ ĥ(n)φ N )], because what gves a nonzero contrbuton from the antsymmetrzer  = (N!) 1 (1 + other permutatons) s only the frst term wth the operator of multplcaton by 1. Other terms dsappear after any attempt of ntegraton. As a result, we have F = φ 1 ĥ φ 1 + φ ĥ φ + φ N ĥ φ N = h, (M.9) whch we wanted to show. Operator Ĝ Now let us consder the expresson for G: G = N! Â(φ 1 φ φ N ) Ĝ  φ 1 φ φ N ),

6 e114 Appendx M where once agan N! comes from the normalzaton of ψ. Takng (smlarly as above) nto account that  = Â,  = Â, Ĝ  = ÂĜ,weget G = N! (φ 1 φ φ N )  [ĝ(1, )φ 1 φ φ N +ĝ(1, 3)φ 1 φ φ N + ] = φ 1 (1)φ () ĝ(1, ) φ 1 (1)φ () φ 1 (1)φ () ĝ(1, ) φ (1)φ 1 () + φ 1 (1)φ 3 (3) ĝ(1, 3) φ 1 (1)φ 3 (3) φ 1 (1)φ 3 (3) ĝ(1, 3) φ 3 (1)φ 1 (3) + (M.10) Ths transformaton needs some explanaton. The factor N! before the ntegral s annhlated by 1/N! comng from the antsymmetrzer. The remander of the antsymmetrzer permutes the electrons n the ket [ĝ(1, )φ 1 φ φ N +ĝ(1, 3)φ 1 φ φ N + ]. In the frst term [wth ĝ(1, )], the ntegrals wth only those permutatons of the electrons 3, 4,...,N wll survve that perfectly match the permutaton φ 1 (1)φ () φ N (N), because otherwse the overlap ntegrals of the spnorbtals (over the coordnates of the electrons, 3,...N) wll make them zero. Ths s why the frst term wll gve rse to only two permutatons that result n nonzero ntegrals: we wll have on the frst two postons φ 1 (1)φ (), and the other one wll have φ 1 ()φ (1). Of course, they wll dffer by sgn, whch s why we have the mnus sgn n the second survvng ntegral. A smlar reasonng may be done for the term wth ĝ(1, 3), as well as other terms. Thus, we have shown that G = ( j j j j ) = 1 ( j j j j ), (M.11) < j, j where the factor 1 N(N 1) takes care of the fact that there s only nterelectronc nteractons g(, j) (the upper trangle of the table N N). There s no restrcton n the summaton over, j = 1,,...N, because any attempt of takng the llegal self-nteracton (correspondng to = j) gves zero due to the dentty of the Coulomb ( j j ) and exchange ( j j ) ntegrals. Ths s the formula that we wanted to prove. Specal Case: Double Occupaton The ntegrals n the expressons for F and G contan spnorbtals, and the ntegraton goes over the electronc space-and-spn coordnates. When the spnorbtals are expressed by the orbtals and the spn functons, we may perform the summaton over spn coordnates. The most popular and the most mportant s the double occupaton case, when every orbtal s used to form two spnorbtals 7 : φ 1 (1) = ϕ 1 (1)α(1) 7 The functons here are wrtten as f they were dependent on the coordnates of electron 1. The reason s that we want to stress that they all are one-electron functons. The electron 1 serves here as an example (and when needed may be replaced by other electron). The symbol 1 means (x 1, y 1, z 1,σ 1 ) f t s an argument of a spnorbtal, (x 1, y 1, z 1 ) f t corresponds to an orbtal, and σ 1 f t corresponds to a spn functon.

7 Slater-Condon Rules e115 or φ (1) = ϕ 1 (1)β(1) φ 3 (1) = ϕ (1)α(1) φ 4 (1) = ϕ (1)β(1) (M.1)... (M.13) φ 1 (1) = ϕ (1)α(1) φ (1) = ϕ (1)β(1), (M.14) where = 1,,...,N/. Thus, the one electron spnorbtals that represent the buldng blocks of the Slater determnant are products of a spatal functon (orbtal ϕ) and one of the two smple functons of the spn coordnate σ (α or β functons, cf. p. 400). The frst Slater-Condon rule [Eq. (M.9)] may be transformed as follows (for the defnton of the ntegrals, see p. 399): N F = ĥ = σ ĥ σ = ( ĥ ) h, =1 =1 σ (M.15) where the summatons denoted by MO go over the occuped orbtals (ther number beng N/), the factor results from the summaton over σ, whch gves the same result for the two values of σ (because of the double occupaton of the orbtals). Let us make a smlar operaton wth G. The formula for G s composed of two parts: G = I II. (M.16) The frst part reads as I = 1 σ, j σ j σ, j σ j, σ j σ j where σ,...etc. stands for the spnorbtal composed of the orbtal ϕ and a spn functon that depends on σ. For any par of the values of σ,σ j, the ntegral yelds the same value (at a gven par of, j) and therefore (cf., p. 399) I = 1 4(j j) = (j j). j The fate of part II wll be a lttle dfferent: II = 1 σ, j σ j j σ j, σ = 1 (j j) = (j j), σ j σ j j j j

8 e116 Appendx M because ths tme the summaton over σ and σ j gves the nonzero result n half of the cases when compared to the prevous case. The pars (σ,σ j ) = ( 1, 1 ), ( 1, 1 ) gve a nonzero (and the same) result, whle ( 1, 1 ), ( 1, 1 ) end up wth zero (recall that by conventon n thentegral,theelectronshavetheorder11).fnally,thedouble occupaton leads to MO G = [(j j) (j j)]., j (M.17) II Slater-Condon Rule Suppose that we are nterested n two matrx elements: F 1 ψ 1 ˆF ψ and G 1 ψ 1 Ĝ ψ and the two Slater determnants ψ 1 and ψ dffer only by that spnorbtal φ n ψ 1 has been replaced by φ (orthogonal to all other spnorbtals)nψ. Then the Slater-Condon rule states that F 1 = ĥ (M.18) G 1 = j=1 ( j j j j ) (M.19) Proof : Operator ˆF Usng ˆF  =  ˆF,  = Â, andâ = Â, we obtan  ˆF  =  ˆF  =  ˆF =  ˆF, and therefore F 1 = N! φ 1 φ  ˆF φ 1,...φ...φ N. F 1 = N! φ 1 φ φ φ N  [ĥ(1)φ 1,...φ φ N + φ 1 ĥ()φ φ φ N + φ 1,...φ ĥ(n)φ N ] = ( 1) p φ 1 φ φ φ N P ˆP[ĥ(1)φ 1,...φ φ N + φ 1 ĥ()φ φ φ N + φ 1,...φ ĥ(n)φ N ]. Note frst that the only ntegral to survve should nvolve φ and φ n such a way that t leads to the one-electron ntegral φ ĥ φ. Ths however happens only f the th term n the square bracket ntervenes [that wth ĥ()]. Indeed, let us take an ntegral that s not lke that ( = 1): φ 1 φ φ φ N ˆPĥ(1)φ 1 φ φ φ N. Whatever permutaton ˆP s, ĥ wll always go wth φ 1, whle φ wll be therefore wthout ĥ. When ntegratng over the electronc coordnates, we

9 Slater-Condon Rules e117 obtan a product of one-electron ntegrals (for subsequent electrons), and n ths product, one always pnponts the overlap ntegral of φ multpled by one of the spnorbtals φ 1,φ,..,.φ N. Ths ntegral (and therefore the whole product) s equal to 0 because φ s orthogonal to all the spnorbtals. An dentcal reasonng can be gven for ĥ(), ĥ(3),..., but not for ĥ(), andwe obtan: F 1 = P ( 1)p φ 1 φ φ φ N ˆP[φ 1 φ ĥ()φ φ N ]. The only ntegral to survve s that whch corresponds to ˆP = 1, because n other cases, the orthogonalty of the spnorbtals wll make the product of the one-electron ntegrals equal zero. Thus, fnally, we prove that F 1 = h. (M.0) Operator Ĝ From  = Â, ÂĜ  = ÂÂĜ = ÂĜ, we obtan the followng transformaton: G 1 = N! Â(φ 1 φ φ N ) ÂĜ φ 1,...φ φ N = N! Â(φ 1 φ φ N ) {[ĝ(1, ) φ 1,...φ φ N ] + [ĝ(1, 3) φ 1,...φ φ N ]+ } = 1 ( 1) p ˆP ( ) φ 1 φ φ N ĝ(k, l) φ1,...φ φ N. k,l P The number of g terms s equal to the number of the nterelectronc nteractons. The prme n the summaton k, l = 1,,...,N over nteractons ĝ(k, l) means that k = l (we count the nteractons twce, but the factor 1 takes care of that). Note that, due to the orthogonalty of the spnorbtals, for a gven ĝ(k, l), the ntegrals are all zero f k = and l =. Thus, the ntegrals to survve have to have k = or l =. Therefore (prme n the summaton means the summaton ndex to be excluded), G 1 = 1 = 1 = + 1 ( 1) p ˆP ( ) φ 1 φ φ N ĝ(, l) φ1,...φ φ N l P ( 1) p ˆP ( ) φ 1 φ φ N ĝ(k, ) φ1,...φ φ N k P [ φ φ l φ φ l φ φ l φ l φ ] + 1 l [ φ φ j φ φ j φ φ j φ j φ ], j [ φ φ k φ φ k φ φ k φ k φ ] because only those two-electron ntegrals wll survve that nvolve both φ and φ, and two other spnorbtals nvolved are bound to be dentcal (and have ether the ndex k or l dependng on whether l = or k = ). The dfference n the square brackets results from two successful permutatons ˆP n whch we have the order, j or j, (n the last term). Fnally, leavng for the k

10 e118 Appendx M sake of smplcty only the ndces for the spnorbtals, we obtan G 1 = j( =)[ j j j j ], (M.1) and after addng 0 =,wehave 8 G 1 = j { j j j j }. (M.) Ths s what had to be demonstrated. III Slater-Condon Rule If ψ 1 and ψ dffer by two spnorbtals, say, n ψ 1 are φ and φ s,andnψ,wehaveφ and φ s (normalzed and orthogonal to themselves and to all other spnorbtals).e., φ replaces φ whle φ s replaces φ s (all other spnorbtals are n the same order) then F 1 = 0 G 1 = s s s s (M.3) (M.4) Proof : Operator ˆF F 1 = N! (φ 1 φ φ N ) Â ˆF(φ 1 φ φ N ) = N! (φ 1 φ φ N ) Â{(ĥ(1)φ 1 φ φ N ) + (φ 1ĥ()φ φ N ) + (φ 1 φ ĥ(n)φ N )} = 0, where the spnorbtals n ψ have been labeled addtonally by prmes (to stress they may dffer from those of ψ 1 ). In each term, there wll be N 1 overlap ntegrals between spnorbtals and one ntegral nvolvng the ĥ. Therefore, there always wll be at least one overlap ntegral nvolvng dfferent spnorbtals. Ths wll produce zero. 8 Wth ths formula, we may forget that the ntegraton has been carred out over the coordnates of the electrons and j. It does not matter what s the symbol of the coordnate over whch an ntegraton s performed n a defnte ntegral. When n the future we wll have to declare whch coordnates we are gong to ntegrate over n j j, then absolutely safely we can put any electrons, n the present book t wll be electron 1 and electron.

11 Slater-Condon Rules e119 Fg. M.1. Four Slater-Condon rules (I,II,III,IV) for easy reference. On the left sde, we see pctoral representatons of matrx elements of the total Hamltonan Ĥ. The squares nsde the brackets represent the Slater determnants. Vertcal lnes n bra stand for those spnorbtals, whch are dfferent n bra and n ket functons. On the rght sde, we have two square matrces collectng the h j and j j j j for, j = 1,...,N. The gray crcles n the matrces symbolze nonzero elements.

12 e10 Appendx M Operator Ĝ Somethng wll survve n G 1. Usng the prevous arguments, we have G 1 = N! (φ 1 φ φ N ) Â(g(1, )φ 1 φ φ N ) + (g(1, 3)φ 1 φ φ N ) + = φ 1 φ g(1, ) φ 1 φ φ 1φ g(1, ) φ φ 1 + φ 1 φ 3 g(1, 3) φ 1 φ 3 φ 1φ 3 g(1, 3) φ 3 φ 1 + = φ 1 φ φ 1 φ φ 1φ φ φ 1 + φ 1 φ 3 φ 1 φ 3 φ 1φ 3 φ 3 φ 1 + Note that N! cancels 1/N! from the antsymmetrzer, and n the ket we have all possble permutatons. The only term to survve has to engage all four spnorbtals:,, s, s ; otherwse, the overlap ntegrals wll kll t. Therefore, only two terms wll survve and gve G 1 = s s s s. (M.5) IV Slater-Condon Rule Usng the above technque, t s easy to show that f the Slater determnants ψ 1 and ψ dffer by more than two (orthogonal) spnorbtals, then F 1 = 0andG 1 = 0. Ths happens because the operators ˆF and Ĝ represent a sum of at most two-electron operators, whch wll nvolve at most four spnorbtals and there wll be always an extra overlap ntegral over orthogonal spnorbtals. 9 The Slater-Condon rules are schematcally depcted n Fg. M.1. 9 If the operators were more than two-partcle ones, then the result would be dfferent.