HE ATOM & APPROXIMATION METHODS MORE GENERAL VARIATIONAL TREATMENT. Examples:

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1 5.6 4 Lecture #3-4 page HE ATOM & APPROXIMATION METHODS MORE GENERAL VARIATIONAL TREATMENT Do t restrict the wavefuctio to a sigle term! Could be a liear combiatio of several wavefuctios e.g. two terms: χ = c χ + c χ For simplicity we ll cosider: χ, χ orthoormal, o variatioal parameters c, c real, variatioal parameters Examples: -D oscillator with cubic, quartic potetial terms: x could be HO groud ad excited state wavefuctios χ ( x ), χ ( ) He atom -electro problem χ ( r, r, r r ) χ (, ) could be product H atom groud ad excited state wavefuctios χ (r r r φ ( r, ) = φ ( ) ( r r ) s ( ) ) ( r ) s s χ, = φ r φ s Best values of c ad c give miimum E var χ ot yet ormalized, so ca't write E var = χ H ˆ χ d τ ˆ τχ χ τ τ ( dτχ χd τ d c χ * * d H d d c + c χ ) Hˆ (c χ + c χ ) E var = = * * τ( χ + cχ )(cχ + c χ ) * ˆ c d τχ H χ + cc dτχ H χ + cc dτχ H χ + c d τχ H χ = * * * * c dτχ χ + cc d τχ χ + c c d τχ χ + c dτχ χ * ˆ * ˆ * ˆ Abbreviate itegrals

2 5.6 4 Lecture #3-4 page ch + cch + cch + ch var cs + ccs + ccs + cs E = I our case S ij = δ ij but we ll keep these terms for geerality Also ofte H = H but we ll keep both terms for ow ( c S + cc S + cc S + c S ) E var = c H + cc H + cc H + c H Miimize E : Differetiate w.r.t. parameters c c var, ( var cs + ccs + ccs + cs + cs E + c( S + S ) E = ch + ch + c ) E c var var H (c S + ccs + ccs + cs ) + c S + S ) E + c S E = ch + ch + c c E var ( H E var var For best c, c, = E = c c c ( H S E var ) + c H + H ( S + S ) E var = c H + H (S + S ) E var + c (H S E var ) = Drop subscript o E var below ( ) ( ) c H S E H + H S + S E = H + H S + S E (H S E c ( ) ) Solutios give by the secular determiat (H S E ) H + H (S + S ) E H + H (S + S ) E (H S E) = All this follows from χ = c χ + c χ & miimizatio of the eergy No assumptios made about the coefficiets or the itegrals

3 5.6 4 Lecture #3-4 page 3 wheever we costruct a wavefuctio from two other fuctios, ψ = c φ + c φ we ca immediately solve for the eergies by solvig the secular determiat with the itegrals as defied. We ca also solve for the costats c, c for each eergy, which gives the wavefuctios. Ofte, ad i our case, S = S ad H = H (ote H is Hermitia) Secular determiat becomes ( H S E ) ( H S E ) ( H S E ) ( H S E ) = Also ofte, S = S = i which case (H S E ) H H ( H S E ) = ad also ofte, S = S = i which case (H E H ( H E) H = H ( H E) )( E) H = E ( H + H ) E + H H H = H + H ± ( H + H ) 4( H H H ) H + H H H E = = ± + H More geerally if we write wavefuctio as liear combiatio of N fuctios, ψ = c φ + c φ c N φ N aalogous treatmet yields N eergies solved through secular determiat

4 5.6 4 Lecture #3-4 page 4 H S E H S E L H S E N N H S E H S E L H S E N N M M M H N S N E H N S N E L H NN S NN E = We ll see this i costructig MO s out of AO s, for example. HE ATOM PERTURBATION THEORY ˆ H = ( + ) Z + + H ˆ ( ) + H ˆ ( ) He r r r Z Z, ) = separable i r ad r Ĥ ( ) (r r r r ˆ H ( ) ψ ( ) ( ) = E ( ) ψ exactly soluble part He atom groud state -electro wavefuctios ψ ( ) (, ) s ( ) ( ) r r = φ r φ s r = Z 3 Z 3 e Zr Zr e i atomic uits π π E ( ) Z = Z = = E s + E s = Z Z = 4 a.u. = 8.84 ev E expt =.93 a.u. = 79. ev Neglect of electro-electro repulsios is iadequate PERTURBATION THEORY ( ST order) H ˆ = H ˆ ( ) + Hˆ () H ˆ ( ) ψ ( ) = E ( ) ψ ( ) Expad ψ ad E: ( ) ψ =ψ +ψ () ( ) () E = E + E ( Hˆ ( ) + H ˆ () )(ψ ( ) +ψ ( ) ( ) = (E ) + E ( ) ( ) )(ψ +ψ ( ) )

5 5.6 4 Lecture #3-4 page 5 ( ) ( ) () ( ) ( ) () () () ( ) ( ) () ( ) ( ) () () () Ĥ ψ + Ĥ ψ + Ĥ ψ + Ĥ ψ = E ψ + E ψ + E ψ + E ψ First terms o each side are equal. Last terms o each side are egligible. ˆ ( ) ( ) ˆ ( ) ( ) ( ) ( ) ( ) ( ) H ψ + H ψ = E ψ + E ψ () Solve for ψ ( * ) ad E (). Left-multiply by ψ ad itegrate: ( ) ˆ ( ) ( ) ( ) ˆ dτψ * ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) H ψ + dτψ * H ψ = dτψ * E ψ + dτψ * E ψ ( * ) ˆ ( ) ( ) ˆ ( ) ( ) () ˆ ( ) Note dτψ H ψ = dτ (H ψ ) * ψ sice H is a Hermitia operator ( ) ( ) ( * ) ( * ) ˆ ( ) ( ) ( ) ( * ) ( ) ( ) dτψ E ψ + dτψ H ψ = E dτψ ψ + E () ( )* () ( ) E = dτψ Ĥ ψ First-order correctio to eergy E ( ). HELIUM ATOM: PERTURBATION THEORY TREATMENT 3 ( ) Z 3 ψ, Z Zr (r r ) = φ s ( r ) φ s ( r ) = e π π Ĥ () = r e Zr () ( ) Z 6 τ Z 6 e Zr * e Zr ˆ Zr Zr Zr 5 Zr dτ = Z (McQ-S. prob. 7-3) π r π r 8 () ( ) E = dτψ H ψ = d e e e e = 5 E = E ( ) + E ( ) = Z + Z = a.u. = ev 8 4 SUMMARY:

6 5.6 4 Lecture #3-4 page 6 Neglect of e ( ) - e repulsio E = 4 a.u. = 8.84 ev st order perturbatio theory E ( ) + E ( ) = 4 a.u. = ev 3 3 th order perturbatio theory E ( ) + E ( ) E ( ) =.94 a.u. = 79. ev Experimet E =.93 a.u. = 79. ev Perturbatio theory results may be greater or less tha the true eergy, ulike variatioal results. Perturbatio theory ca be used to calculate ay eergy level, ot just the groud state. PERTURBATION THEORY (cot d) Now retur to ˆ ( ) ( ) ˆ ( ) ( ) ( ) ( ) ( ) ( ) H ψ + H ψ = E ψ + E ψ () Solve for ψ by writig ψ as liear comb. of uperturbed wavefuctios: ψ =ψ ( ) ( ) ( ) () +ψ =ψ + a j ψ j j ( * ) Left-multiply by ψ k ad itegrate: dτψ k H ψ + dτψ k ( * ) ˆ ( ) ( ) ( * ) ( ) ( Ĥ ψ ) * ( ) ( ) ( ) ( ) ( ) = dτψ ( ) k E ψ + dτψ * k E ψ

7 5.6 4 Lecture #3-4 page 7 ( ) k j * dτψ Hˆ ( ) a ψ () + dτψ ( * ) Hˆ ( ) ψ ( ) = dτψ ( * ) E ( ) a ψ + dτψ * E ψ j j k k j () ( ) ( ) ( ) j j k () ( * ) () ( * ) ˆ ( ) ( ) ( ) ( * ) () ( ) * aj E j dτψ k ψ j + dτψ k H ψ = E aj d τψ k ψ j + E dτψ k ψ j j ( ) ( ) () () ( ) ( )* ( ) ( )* ( ) ( ) a (E E ) = E dτψ ψ dτψ Ĥ ψ k k k k () k E ( = = dτψ * ) ˆ ( ) ( ) H ψ ( ) ˆ ( ) ( ) () dτψ * k H ψ Ĥ k k a k = () () () E E k E E k () () ( * ) Hˆ ( ) ( ψ ) Ĥ j ψ () =ψ ( ) + dτψ j j j j j ψ =ψ ( ) +ψ ( ) =ψ ( ) + E () E () j E () E () j First-order correctio to the wavefuctio ψ () (See also McQ. prob. 7-4) Secod-order correctio to the eergy: (See also McQ. prob. 7-4) Expad ψ ad E: ( ) () () ψ =ψ +ψ + ψ ( E = E ) () + E + E () (Ĥ ( ) + Ĥ ( ) )(ψ ( ) +ψ () + ψ () ) = (E ( ) + E () + E () )(ψ ( ) +ψ () + ψ () ) ( ) ( ) ( ) ( ) Ĥ ψ = E ψ zero order () ( ) ( ) () () ( ) ( ) () Ĥ ψ + Ĥ ψ = E ψ + E ψ first order Collect secod-order terms ( ) () () () () Ĥ ψ () + Ĥ ψ = E ( ) ψ () + E ψ + E () ψ ( ) ( * ) Left-multiply by ψ ad itegrate:

8 5.6 4 Lecture #3-4 page 8 ( ) ˆ ( ) () ( ) ˆ dτψ * ( ) ( ) H ψ + dτψ * H ψ ( * ) ( ) () ( * ) () () ( * ) () ( ) = dτψ E ψ + dτψ E ψ + dτψ E ψ ( )* () () () ( )* () () ( )* () ( )* () dτψ E ψ = E dτψ ψ = E dτψ a ψ = a dτψ ψ = dτψ j j j j ( ) ( ) () ( ) ( ) () ( ) ( ) () * Ĥ ψ = dτ (Ĥ ψ ) * ψ = dτψ * E ψ j j E = dτψ * Ĥ ψ Ĥ () Ĥ () Ĥ () = dτψ * Ĥ j ψ j j () () j = () j E E j j E E j () ( ) () () ( ) () () () PERTURBATION THEORY EXAMPLE: HO + liear potetial term Ĥ () = h d + kx µ dx Ĥ () = λh ωα x ψ () = α 4 e α x ψ () α 4 = α α x α = kµ ( x) e π π h () k E = + hω ω = µ FIRST ORDER: Fid E () for lowest HO levels α () ()* ˆ () () α x E = ψ H ψ dx = λh ωα xe dx = π () ()* ˆ () () α 4α 3 α x E = ψ H ψ dx = λh ωα ( ) x e dx = π It s as log as ψ ()* ψ () = ψ () is of eve symmetry! Simple model for polar molecule i a electric field HO with particle of charge q i field E (alog x directio) Ĥ () = U E ( x )= qe x () No effect to first order sice ψ are symmetric about x = Same for homouclear diatomic molecule: charge desity symmetric about x = Same for ANY molecule with iversio symmetry, e.g. SF 6 etc.!

9 5.6 4 Lecture #3-4 page 9 But ot for heterouclear diatomics or ay molecule without iversio symmetry! () The eergy levels chage to first order i the field: E E How much chage depeds o molecular dipole momet µ Ĥ () = U ( )= ( q x = q x E x i E i ) E i i i i µe µ= if charges are distributed symmetrically about x = More geerally Ĥ () = µ E vector quatities d ORDER: α 4 Recall geeral HO solutios: ψ () = NH (α α x xe ) N = (!) π ad recursio relatio for the Hermite polyomials H : ξ H ξ = H + ( ξ ) + H ( ξ ) ( ) Easy to verify for the first few: H ξ = H ξ = ξ H ξ = 4ξ H ξ = 8ξ ξ ( ) ( ) ( ) 3 ( ) 3 Fid E () for lowest HO levels Hˆ () Hˆ () () j j E = () E j j E () ˆ () ()* ˆ () () ()* () ()* N h () H j = ψ j H ψ dx = λ ω ψ j (α x)ψ dx = λh ω ψ j ψ dx N λh ω ()* () = ψ ψ dx j Oly j = term i the sum is ozero! Ĥ () = λhω = Ĥ () ˆ () ˆ () H H λ ( hω ) E () = = = λ h ω () () E E hω

10 5.6 4 Lecture #3-4 page Similar for ext level E () = Hˆ () Hˆ () j j () () j E E j ˆ () ()* ˆ () () ()* () ()* N h () N () H j = ψ j H ψ dx = λ ω ψ j (α x)ψ dx = λh ω ψ j ψ + ψ dx N N 8 ()* () ()* () j = λh ω ψ ψ dx + h ψ ψ dx j λ ω Oly j = ad j = terms i the sum are ozero Ĥ () = λhω = Ĥ () Ĥ () = λhω = Ĥ () ( ) ˆ () ˆ () ˆ () ˆ () λ hω λ hω E () = + = + = λ h ω () () () () E E E E hω hω H H H H ( ) same correctio! Actually HO + liear potetial problem ca be solved exactly! U = kx + h α λ ω x = kx + λh ωα x + λ ω h where d = λh ωα k h λ ω k ( x d ) λ ω h This is just HO with potetial eergy miimum shifted to x = d ad U = λ h ω (costat). Wavefuctios are the same except cetered aroud x = d istead of x = Eergies are the same except the miimum of eergy (the potetial eergy miimum) has bee set to λ hω istead of. Spacigs betwee eergy levels are still hω. HELIUM ATOM HARTREE-FOCK METHOD He electroic wavefuctio

11 5.6 4 Lecture #3-4 page Writte as product of two H-like s orbital wavefuctios: ~ ~ r r φ Z Z χ (, ) = s ( r ) φ s ( r ) where ~ 3 ~ Z Z ~ s r = π a φ ( ) e Zr a -electro H atom orbitals How do we kow the s orbitals are the best choices? WE DON T! MORE GENERAL APPROACH: Do t specify or restrict the form of the -electro wavefuctios! χ (r, r ) = φ ( r ) φ ( r ) Problem: Fid the best -electro orbitals, labeled the Hartree-Fock orbitals φ HF ( r ) φ HF, ( r ) ad solve for the eergy E var = E HF Iterpret φ * ( r ) φ ( r ) = φ ( ) as charge desity for electro #. r The effective potetial eergy for e- # at poit r due to e- # is φ ( )dr U eff ( r ) = φ * ( r ) r r Coulomb potetial felt by e- # averaged over all positios of e- #, weighted by φ ( r ) Now write effective Hamiltoia ad Schrodiger eq. for just electro #! Recall full Hamiltoia

12 5.6 4 Lecture #3-4 page ˆ H He (r r ( + r r r, ) = + ) Z + ad write ˆ eff Z H ( r ) = + r eff Ĥ ( r ) φ ( r ) = εφ ( r ) U eff ( ) r Hartree-Fock equatio for the He atom i just oe electro s coordiates! But to solve it, we eed φ ( r )! HOW TO DO IT: THE SELF-CONSISTENT FIELD (SCF) METHOD. Start with trial wavefuctio ( ) φ r eff ( ) ( r ). Use φ r to calculate U U φ r d eff ( r ) = φ * ( r ) ( ) r r 3. Isert U eff ( ) Ĥ eff r ito ( r ) ad solve HF equatio for φ ( r ) ˆ eff Z eff r r H ( r ) = + U ( ) ˆ r φ r = εφ ( r ) H eff ( ) ( ) eff ( ) ( r ). Use φ r calculate U φ r r U eff ( r ) = φ * ( ) d r ( ) r 3. Isert U eff eff ( r ) ( r ) ( ) ito Ĥ ˆ eff Z H ( r ) = + r Ĥ eff ( r ) φ ( r ) = ε φ ( r ) U eff ( ) r ad solve HF equatio for φ r

13 5.6 4 Lecture #3-4 page3 eff ( ) ( r ). Use φ r calculate U 3. Isert U eff ( ) ito Ĥ eff r ( ) r ( ) ad solve HF equatio for φ r etc. Cotiue util SELF-CONSISTENT results φ HF ( r,φ HF ) ( r ) are reached The χ (r r, ) = φ HF ( r ) φ HF ( r ) Calculate the Hartree-Fock eergy usig the full Hamiltoia: E HF HF = * ( ) φ HF* ( r ) ) + + HF Z φ r φ HF ( r ) d d r r r I + I + J φ r ( + ( ) r r where I = φ HF * ( ) r r φ HF* ( ) Z φ HF ( r ) dr I Z φ HF = r ( r )dr = I r HF J = * HF r * r φ HF r φ HF φ ( ) φ ( ) ( ) ( r ) drd r r Coulomb Itegral Result for the Helium atom: E HF = ev E expt = -79. ev COMPARISON OF ENERGIES:

14 5.6 4 Lecture #3-4 page4 * ( ) ˆ eff ( ) ( ) ε = φ r H r φ r dr = I + J = E I ε = I + J ε + ε = I + I + J = E + J E What do these terms represet physically? I looks like the eergy of the Helium io, calculated usig the HF orbitals. (which were determied usig the Coulomb iteractio betwee electros) The ε = total eergy of He atom - eergy of He io = ioizatio eergy! E io ε HF value: Exp t:.98 au = 5. ev.94 au = 4.6 ev Mai approximatio: use of same orbitals for the atom ad the io. For io, should use H-atom orbital with Z= o iterelectro repulsio.