Adiabatic (Born-Oppenheimer) approximation

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Geraldine Hodges
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1 Chpter 1 Adibtic (Born-Oppenheimer) pproximtion First write our the Hmiltonin for the nuclei-electron systems. H = 1 A A M A + 1 A B Z A Z B R AB i,a Z A 1 i + r ia m i e i,j 1 r ij (1.0.1) We im t seprting this problem into n electron problem prmetric in the nuclei coordintes nd nuclei prt. First prt of B-O pprox: Write the totl wvefunction s product of nucler prt ν(r) nd n electronic prt Ψ({r}; {R}), where the semicolon represents the fct tht the nucler coordintes re sttic prmeters. Φ({r}, {R}) == ν({r})ψ({r}; {R}) (1.0.) Second prt of B-O pprox: determine Ψ({r}; {R}) from the prt of the Hmiltonin with sttic nuclei nd ignore cross terms of the nucler kinetic opertor with the electronic prt of the wvefunction. H Φ({r}, {R}) >= A A M A + 1 A B Z A Z B + H el (r; R) Ψ({r}; {R}) > ν({r}) > R AB (1.0.3) = H el (r; R) Ψ({r}; {R}) > + Ψ({r}; {R}) > 1 A + 1 M A A A A B Z A Z B R AB ν({r}) > 1 [ ν({r}) > M A Ψ({r}; {R}) > + A ν({r}) > A Ψ({r}; {R}) > ] A = H el (r; R) Ψ({r}; {R}) > + Ψ({r}; {R}) > 1 A + 1 M A A A B Z A Z B R AB ν({r}) > (1.0.3b) (1.0.3c) H el = i,a Z A 1 i + r ia m i e i,j 1 r ij (1.0.3d) In words: we wnt to solve the electronic problem for sttic nuclei. This will then crete n effective potentil for the nuclei. This effective potentil is referred to s the dibtic (Born-Oppenheimer) surfce. I m now going to drop lbels of the set of nucler coordinte {R} to which prticulr multi-electron wvefunction refers. 1

2 Chpter Hrtree Fock Method.1 Slter Determinnts Eqution 1.0.3d is still multi-body problem. In order to proceed, we must mke n ssumption on the nture of the electronic wvefunction Ψ({r}). The simplest ssumption is tht the wvefunction is composed of product of one electron wvefunctions: Ψ({r}) = χ 1 (r1)χ (r)χ 3 (r3) > (.1.1) This is known s Hrtree product. The problem with eqution.1.1 is tht it gives electrons n identity. In other words, if I swp two electrons I get wvefunction with potentilly different properties. In still other words, the wvefunction is not fermionic wvefunction s it is not ntisymmetric under exchnge of electrons. Therefore the simplest physicl wvefunction hs form: Ψ({r}) = n=n! 1 (( 1) pn P n (χ 1 (r1)χ (r)... )) (.1.) N! n where P n represents the permuttions opertor for electron coordintes nd p n represents the prity (even-ness) of the number of permuttions. A more convenient wy of representing the sme eqution is: χ 1 (r1) χ (r1) χ n (r1) 1 χ 1 (r) χ (r) χ (r) Ψ({r}) = N!.... (.1.3) χ n (rn) χ n (rn) To cut down on the suffix-mdness such Slter determinnt will be referred to s: χ 1 χ χ 3...χ n > (.1.4) The Hrtree Fock method consists in finding the vritionl vlue of the electronic energy for such wvefunction.. Mnipulting Slter Determinnts Most of the efforts in quntum chemistry re concerned with finding expecttion vlues between Slter determinnts nd the Hmiltonin from eqution 1.0.3d. There re two prts in tht Hmiltonin, one which dependends on one electron only O 1 = i h(ri) = Z A i,a r ia 1 i i m e nd one which depends on pirs of electrons: O = 1 i,j r ij. Let us consider wht the expecttion vlue of O 1 is. < Ψ({r}) O 1 Ψ({r}) >=< Ψ({r}) i h(ri) Ψ({r}) > (..1)

3 = N < Ψ({r}) h(r1) Ψ({r}) > (..1b) This mnipultion is possible becuse ll electron coordintes re indistinguishble from ech other (thnk you nti-symmetry!). Ech Slter determinnt is composed of N! permuttions of spin orbitls χ n (rj). When we re integrting over the vribles r, r3, r4 etc. those vrible must pper in the sme order in both wvefunctions. This is becuse of the orthonormlity of the one electron spin orbitls. In other words: drχ i (r)χ j (r) = δ ij (..) For prticulr spin orbitl χ(r1) ssocited with r1 there will be (N 1)! such combintions. (N 1)! combintions integrte to 1. Hence: All < Ψ({r}) O 1 Ψ({r}) >= i N(N 1)!(N!) 1 (< χ i (r1) h(r1) χ i (r1) >) (..3) In proxysm of lziness, this eqution is further simplified by nother piece of nottion: we drop the electron coordintes lltogether, nd drop the χ: < Ψ({r}) O 1 Ψ({r}) >= i (..4) Wht bout O? In this cse the opertors involve lwys pirs of electrons: Agin, ech coordinte is undistiguishble, so we cn write: O = r 1 + r 13 + r (..5) So: = N(N 1) (N!) 1 m O = N(N 1) r 1 (..6) ( 1) pm+pn P m (χ 1(r1)χ (r)... ) 1 P n (χ 1 (r1)χ (r)... ) (..7) r 1 n m Given tht the spin orbitls χ m nd χ n re ssocited with coordintes r1 nd r, orthonormlity of the spin orbitls mens there re only (N )! permuttions of the other integrtion vribles tht integrte to 1. Therefore: = 1 m n m dr1drχ m(r1)χ n(r) 1 r 1 (χ m (r1)χ n (r) χ n (r1)χ m (r)) (..8) Agin, this expression is pin to write out, so we use some nottion two re-express it: < ij kl >= dr1drχ i (r1)χ j (r) 1 χ k(r1)χ l (r) (..9) Also: < ij kl >=< ij kl > < ij lk > (..10) So with this beutiful nottion we cn concisely write the expecttion vlue of single Slter determinnt with the Hmiltonin: < Ψ({r}) H Ψ({r}) >= i + 1 < mn mn > (..11) Minor point: since < mm mm >= 0, we could drop the n m condition. NB: All the integrtions over r still contins integrtion over the spin coordintes!!! m n m 3

4 .3 Integrting out spin Let us think up n exmple: the energy of Lithium tom in the ground stte. Let us immgine two electrons occupy sptil orbitl 1 > nd the third is in some different orbitl >, I shll express spin by br. So therefore our wvefunction is: Ψ({r}) = 1 1 > (.3.1) Spin does not mtter one hoot when clculting the one electron prt: spin simply integrtes out! Therefore: wht bout the two electron prt? < Ψ({r}) O 1 Ψ({r}) >= < 1 h 1 > + < h > (.3.) < Ψ({r}) O Ψ({r}) >= 1 ( (.3.3) < > < > + < 1 1 > < 1 1 > + < > < > + < 1 1 > < 1 1 > < 1 1 > < 1 1 > + < 1 1 > < 1 1 >) = 0.5(< > + < 1 1 > < 1 1 > + < > + < 1 1 > < 1 1 > < 1 1 > + < 1 1 >) (.3.4) Let us introduce nother two pieces of shorthnd: coulomb nd exchnge integrls. The Coulomb integrl: J b = (.3.5) The exhnge integrl: K b = (.3.6) Using these definitions we cn write the energy of the Lithium tom s: = < 1 h 1 > + < h > +J 11 K 11 + J 1 K 1 (.3.7) this is becuse K 11 == J 11. So in other words, the Coulomb opertor opertes over ll pirs of electrons, irrespective of spin wheres the Exchnge opertor opertes only for pirs of equl spin. It lso seems tht exchnge voids self interction of electrons by the Coulomb opertor..4 Vritionl solution of the Hrtree Fock problem We wnt to minimise the expression: E 0 [{χ}] = i + 1 < mn mn > (.4.1) m n with respect to the one electron spin orbitls χ 1 χ... given the condition: dr1χ (r1)χ b (r1) = [ b] = δ b (.4.) Let us construct the vritionl functionl: L[{χ}] = E 0 [{χ}] ɛ b ([ b] δ b ) (.4.3) b 4

5 Let us mke smll vrition of the orbitls {δχ}. The corresponding vrition in the functionl L must be zero hence: δe 0 [{χ}] = ɛ b δ[ b] (.4.4) Let us evlute the left hnd side of this eqution: δe 0 [{χ}] = < (δ) h > + < (δ)b b > < (δ)b b > +c.c. (.4.5) b NOTE: the fctor of hlf hs disppered! We wnt to extricte the (δ) terms from this eqution. In order to do so, let us define two one electron opertors: the Coulomb nd Exchnge opertors: J (r1) b >= dr1χ (r1)χ (r1)χ b (r) (.4.6) b The exchnge opertor is wierder: K (r1) b >= dr1χ (r1)χ b(r1)χ (r) (.4.7) Both opertors depend on the electron wvefunction, mking them wierd 1 electron opertors. The exchnge opertor ctully depends on the spin orbitls it opertes on nd is therefore non-locl! The Coulomb opertor J is clerly expressing the men coulomb ttrction from n electron in spin-orbitl χ. Exchnge cnnot be so clerly understood: it is present only becuse of the nti-symmetric nture of the wvefunction. It cncels out the Coulomb ttrction of spin orbitl with itself (though to me this seems like piece of sleight of hnd, depending on dropping the m n. More importntly, it reduces the repulsion between electrons of prllel spin. Using these convenient devices the eqution for L is written: δl = dr1δχ (r1) ( h(r1)χ (r1) + b (J b (r1) K b (r1))χ (r1) b ɛ b χ b (r1) ) + c.c. (.4.8) Here comes slightly subtle bit: since the vritions re rbitrry, nd re independent of ech others complex conjugte, the integrnd itself must evlute to zero. ( ) h(r1) + b (J b (r1) K b (r1)) χ (r1) = b ɛ b χ b (r1) (.4.9) We re lmost there! The opertor: h(r1) + b (J b(r1) K b (r1)) is simply the fock opertor f(r1). In order to put this eqution in Cnonicl form, we must hve the Lgrnge multipliers ɛ b s digonl mtrix..5 Cnonicl Hrtree Fock equtions Consider n rbitrry unitry trnsformtion U of the spin orbitls. The Fock opertor is unvrint under such trnsformtion. The one electron prt h is clery independent of it, let us consider the exchnge opertor (for the coulomb opertor this is shown in pge 1). = χ b(r1) = U cb χ c (r1) (.5.1) c K (r1)χ b () = dr1 χ (r1)χ b (r1)χ () (.5.1b) K (r1)χ b() = dr1 ( U iχ i (r1))( U jb χ j (r1))χ () (.5.1c) i j dr1 UiU jb (χ i (r1)χ j (r1))χ () = dr1 (χ i (r1)χ i (r1))χ () (.5.1d) ij i 5

6 From the fundmentl property of unitry mtrix. So if we tke eqution.4.9 nd left multiply it by χ c, we get: < χ c f(r1) χ >= ɛ c (.5.) Since this ɛ c is Hermitin mtrix (consequence of the definition of the vritionl functionl), it is lwys possible to define unitry trnsformtion tht digonlises nd leves the fock opertor (nd hence our vritionl energy) intct. So typiclly the cnonicl problem we solve is: f(r1) χ >= ɛ χ > (.5.3) This is rther importnt point in HF: the spin orbitls which we usully cll moleculr orbitls, re not the only orbitls who cn construct Slter determinnt with vritionl energy. They just hppen to be the ones tht digonlise the fock mtrix (defined s F b =< χ f(r1) χ b >). We cn tke ny rottion of those orbitls nd still hve the sme totl energy: nothing is specil bout moleculr orbitls nd they men *lmost* nothing..6 Self Consistent method nd LCAO Eqution.5.3 sys tht if we construct Slter determinnt with spin orbitls which digonlise the Fock opertor, the energy of tht Slter determinnt is the lower thn or equl to ny energy described by Slter determinnt. The ctch is tht the fock opertor itself depends on the spin orbitls! We lso need to find representtion for the 1 electron opertor f, which s we cn see will be chieved by using bsis set. In order to get to prcticl computtionl method to solve this problem it is lso importnt to write down the spin orbitls in term of some bsis set. Without getting into the detil of why pick certin bsis set rther thn nother, let us ssume tht the sptil prt of the spin orbitls hs been written s liner position of some bsis set {φ}. χ >= C i φ i (.6.1) Also we shll need to know the overlp mtrix of this bsis set: < φ j φ i >= S ij (.6.) So let us write integrte the spin degrees of freedom out of the Fock opertor (just ss for the cse if Lithium solved erlier) for closed shell system (one with ll electrons pired in the sme sptil orbitls): =N/ f(r1) = h(r1) + (J (r1) K b (r1)) (.6.3) This cn esily be derived bering in mind tht exchnge interctions exist only between pirs with similr spin. Let us rewrite eqution.5.3 for generl non-orthogonl bsis: f(r1) χ >= f(r1) C i φ i = ɛ Ci φ i (.6.4) now left multiply this eqution by φ j nd integrte: dr1φ j (r1)f(r1) C i φ i (r1) = ɛ Ci φ j (r1)φ i (r1) (.6.5) writing it in mtrix form: FjiC i = ɛ Sj ic i (.6.6) [I should rewrite the previous prt using greek letters for the bsi set but cnnot be bothered]. Let us evlute.6.3 for prticulr bsis set: =N/ F µν =< φ µ h φ ν > + ( < µ ν > < µ ν >) (.6.7) 6

7 remember ll the sptil orbitls: so: F µν =< φ µ h φ ν > + =N/ χ = C λ φ λ (.6.8) (C λ Cσ)( < µσ νλ > < µσ λν >) (.6.9) λσ F µν =< φ µ h φ ν > + λσ P λσ (< µσ νλ > 0.5 < µσ λν >) (.6.10) This is it! We hve written our Fock opertor in prticulr mtrix representtion, using prticulr non-orthogonl bsis set. The Fock mtrix hs one electron prt < φ µ h φ ν > which cn be clculted once t the beginning of the computtion. We first mke guess for the orbitls, then construct the two electron density mtrix P λσ, from this nd the hideous electron integrls < µσ λν > we construct F, solve the generlised eigenvlue problem F C = SCe, plug the solution bck into the density mtrix nd keep going untill subsequent density mtrices re consistent..7 Koopmns Theorem It hs been sid tht the orbitls which digonlise the Fock mtrix re meningless. In fct they hve one cler mening: they re connected to the ioniztion energy of molecule. Consider system with N electrons in N closed shells, then: E0 = < h > + (.7.1) Wht if we remove nd electron from the the Nth orbitl, leving ll other orbitls untouched (frozen orbitl pproximtion)? Wht will the energy of such determinnt be? N b E+ = N < h > + < N h N > + (.7.) + < Nb Nb > < Nb bn > b N b N So: E + E0 =< N h N > +( c < Nc Nc > < Nc cn >) = ɛ N (.7.3).8 Closing Remrks I hve been very sloppy with nomenclture. Ostlund. USE AT OWN PERIL. This is bsiclly rther bd prphrse of Szbo nd 7

8 Bibliogrphy [1] A. Szbo nd N. Ostund, Modern Quntum Chemistry,Dover Publictions 198 8

Chapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1

Chapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1 Tor Kjellsson Stockholm University Chpter 5 5. Strting with the following informtion: R = m r + m r m + m, r = r r we wnt to derive: µ = m m m + m r = R + µ m r, r = R µ m r 3 = µ m R + r, = µ m R r. 4