19 The Born-Oppenheimer Approximation

Transcription

1 9 The Bon-Oppenheme Appoxmaton The full nonelatvstc Hamltonan fo a molecule s gven by (n a.u.) Ĥ = A M A A A, Z A + A + >j j (883) Lets ewte the Hamltonan to emphasze the goal as Ĥ = + A A A, >j j M A AB A A>B (884) Howeve, f we assume that the nuclea moton s so slow elatve to the electons that the nucle can be consdeed fxed wth espect to the moton of the electons. Usng ths we can wte to total wavefuncton as Ψ(, A ) = Φ( A )ψ(, A ) (885) whee ψ(, A ) s a paametc functon of A,.e. fo a fxed nuclea poston. The tme-ndependent Schödnge equaton s then ĤΨ(, A )=EΨ(, A ) (886) o Φ( A )ψ(, A )(887) A, A >j j + A M Φ( A )ψ(, A )=EΦ( A )ψ(, A )(888) A A ealzng that Φ( A )ψ(, A ) = Φ( A ) Iψ(, A ) ] (889) AΦ( A )ψ(, A ) = ψ(, A ) AΦ( A ) ] (890) + a ψ(, A )] A Φ( A )] (89) +Φ( A ) A ψ(, A ) (89)

2 snce and A only woks on electon and nuclea coodnates, espectvely. Insetng ths gves Φ( A ) ψ(, A ) (893) A A, >j j + ψ(, A ) A M Φ( A ) (894) A A a ψ(, A )] A Φ( A )] (895) M A A Φ( A ) M Aψ(, A ) A a (896) = EΦ( A )ψ(, A ) (897) Ignong tems that scale as /M A gves the B.O. appoxmaton ψ(, A )=E el ψ(, A ) (898) A, A >j j and A M Φ( A )=E nuc Φ( A ) (899) A A A M + E el () Φ( A )=E tot Φ( A ) (900) A AB A A>B Often one solve the electonc Schödnge equaton consdeng the nucle as fxed (Ĥel + V NN )φ el = Uψ el =(E el + V NN )ψ el (90) ths gves us a potental enegy suface as a functons of the nuclea coodnates. 3

3 0 The H + molecula-on We seek to solve the electonc Schödnge equaton of the smallest molecule H + fo the nucle fxed at vaous sepaatons. Ĥ el = + (90) A, A >j j The soluton to Schödnge equaton s sepaable f we use ellpsodal coodnates, µ, υ, φ, whee φ s the angle of otaton of the electon aound the ntenuclea axs and the two othe coodnates ae gven by µ = A + B υ = A B The soluton to ths gves a wavefuncton of the fom (903) (904) ψ el = L(µ)M(υ) exp(mφ) (905) π We aleady know the soluton n two lmts. As we get the sepaated atom lmt, and when 0,.e. t becomes He + we get the unted atom lmt. In ode fo us to teat obtan a soluton usng the lnea vaaton method we need to pck a bass. In the sepaated atom lmt an appopate bass set would be s AO s and n the unted atom lmt a hydogen-lke s obtal wth Z=. The sepaated atom lmt bass set consttute a mnmum bass set. The wavefuncton obtaned fom the lnea vaaton method descbes the whole molecule and ae theefoe a one-electon soluton to the molecula Schödnge equaton and ae called molecula obtals: ψ = c A s A + c B s B (906) and ae efeed to as Lnea Combnaton of Atomc Obtal (LCAO). The secula detemnant s then H AA ES AA H AB ES AB H BA ES BA H BB ES BB = 0 (907) We can now use symmety to educe the detemnant to H AA E H AB ES AB H AB ES AB H AA E = 0 (908) 4

4 Expandng the secula detemnant gves and theefoe, the enegy s (H AA E) (H AB ES AB ) = 0 (909) E ± = H AA ± H AB ± S AB (90) Now lets fnd the coeffcents fo each of the oots fom substtutng the E + soluton gves (H AA E)c A +(H AB S AB E)c B = 0 (9) (H AA H AA + H AB H AA + H AB )c A = c B (H AB S AB )(9) +S AB +S AB (H AA + H AA S AB H AA + H AB )C A = c B (H AB + H AB S AB H AA S AB H AB S AB )(93) Theefoe, (H AA S AB H AB )C A = c B (H AB H AA S AB )(94) ψ = c a (s A +s B ) (96) Usng nomalzaton gves c A (s A +s b + s a s b )dv = (97) c A = + SAB (98) Smla fo the negatve oot we fnd c A = c B and the wavefunctons ae ψ + = s A +s B (99) + SAB ψ = s A s B (90) SAB The dffeent ntegals ae gve by S AB = s A s B dv = exp( ) + + /3 ] (9) H AA = s A Ĥ el s A dv = exp( )( + )] (9) H AB = s A Ĥ el s B dv = S AB / exp( )( + ) (93) c A = c B (95) 5

5 At = we get S AB =0.586,H AA = 0.97,H AB = Insetng these nto the enegy solutons gves us E + =.054,E = 0.66 and addng the ntenuclea epulson (/) we get the values of and and smla f we plot the wavefuncton Snce these functons ae ethe symmetc o antsymmetc unde nveson though the mdpont we label them geade and ungeade, espectvely. We know that the angula dependence of the soluton s gven by Φ(φ) = π exp(mφ),m=0, ±, ±, (94) Whch ae egenfunctons of L z. We can theefoe label ou MO s accodng to the value of m. We wll use the Geek lettes, σ, π, δ, to ndcate values of m of 0,,. Snce both of ou solutons ae cylndcally they ae σ obtal, and we denote them σ g and σ u, espectvely. Lookng at the geade soluton we see that s puts electonc chage nto the bondng egon and s theefoe called a bondng obtal. Fo the ungeade functon we see that chage s shfted out of the bondng egon snce t s popotonal to exp( A ) exp( B ), and ae theefoe called an antbondng obtal and ae denoted by σu. 6