Many Electon Atoms The many body poblem cannot be solved analytically. We content ouselves with developing appoximate methods that can yield quite accuate esults (but usually equie a compute). The electons aound a nucleus to a fist appoximation can be consideed independent of one anothe Electons ae indistinguishable, theefoe the wave function must eflect this. Electons have spin of one half; this must be included in the atomic desciption (Pauli Exclusion and spin). Electons can be put into appoximate obitals and the popeties of the many electon systems can be catalogued The Aufbau (building up) pinciple fo placing electons in obitals
Two electons in an Atom Can t solve Sch. Eqn. fo many-electon atoms analytically because of e - -e epulsions. The intenal Hamiltonian fo the two electons aound the nucleus with chage Z is: ˆ ˆ e H + H + ψ, = Eψ, Hˆ e 4 ( ) ( ) 4πε0 = m Ze πε e 0 - x y z e = + + + - tem emoves spheical symmety in He, Z= -
Spin: The Sten-Gelach Result The spin of the electons comes into play in the obital pictue of electons as pat of the desciption of each electon. Spin we found has two possibilities, and comes in units of h-ba like angula momentum. We found the by fom fo angula momentum (although it is not possible to be obital angula momentum) Using the commutato ule of angula momentum we geneated the y component Find that the total spin obeys the ule of angula momentum too. This IS the way to epesent spin. Thee is no othe way. Spin is a type of angula momentum. 0 Sz = 0 0 α = 0 β = 0 Sx = R SzR = 0 0 Sz, Sx = i Sy = 0 0 i S y = i 0 0 Sx = Sy = Sz = ( ) 0 3 0 S = Sx + Sy + Sz = 4 0 3 s =, = s s+ 3 4 ( )
Make up the wave function fom one-electon wave functions (an appoximation) In pinciple, the state of an atom is descibed by a many electon wave function. Howeve, expeience suggests that electons can be teated independently (i.e. valence vs. coe). Can make the obital appoximation. Many electon wave function is poduct of one electon obitals. ψ,,..., = φ φ... φ ( ) ( ) ( ) ( ) n n n Does not mean that electons do not sense each othe. Howeve they will sense each othe in some aveaged way, not tuly diectly coelated (as one might suppose goes on in eal life). 4
The enegy fom appoximate wave functions We cannot solve the eigenvalue poblem but we can compute the enegy of the system if we have an appoximate wave function. E = Ψ Hˆ Ψ dd * elec apx apx If electons ae indistinguishable and we wite the wave function as a poduct of one electon obitals then if we switch the electons twice we must get the same wave function back. Let this be ae intechange opeato: P { ( ) ( )} ( ) ( ) φ φ = p φ φ P p { } { φ ( ) φ ( )} = φ ( ) φ ( ) { } If the system is to be unchanged by the intechange o paity opeato then p = which implies that p =± Theefoe, the wave function is eithe even o odd unde electon 5 exchange. Tuns out it is odd because of spin.
Full symmety Each electon must be in a spin eigenstate because the Hamiltonian does not scamble o mix up the spin states with spatial states. (Late we will note that this does actually happen.) Theefoe the single electon obitals can be a poduct of the spin obital and a spatial obital. The spatial obital can hold two electons. The complete pictue to make sue the wave function changes sign unde paticle exchange is to wite the wave function as a deteminant; called a Slate Deteminant. To do this choose you N obitals, including spin, fo the N electons Put the electon in an obital, put each of the N obitals on the diagonal Fill in the off diagonal elements by moing the obital numbe as you go hoizontally and the electon numbe as you go vetically. ψ =,,..., ( ) n φ α, φ α,..., φn N α N! () () ( ) ( ) ( ) ( ) N 6
Combining Obitals to Keep Spin Eigenfunctions When spin obitals ae combined to make a two electon (o N electon) wave function the new poduct spin states ae not necessaily eigenfunctions of the total spin anymoe. Look at how two spins go togethe to see this poblem and see how to solve it. Notice we left of the facto of h-ba to focus on the pocess. The spin opeatos fo two spins add. (, ) ( ) ( ) S = S + S = S + S S S S z z z z z z z z 0 0 = and = 0 0 0 S 0 z = and Sz = 0-0 Sz (, ) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 = + = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7
Total Spin Do the same combining fo the x and y spins as well. Squae each matix and add them up to get the total S-squaed opeato. In this case (Notice the Total Spin opeato is not even diagonal): The poduct of the spin basis functions is a set of 4 othonomal vectos but they ae not eigenfunctions of the Total Squaed Spin opeato any moe N i= (, ) () S = S i whee q = x, y, z q q S = S S + S S + S S S Total x x y y z z = Total 0 0 0 0 0 0 0 0 0 0 0 α = and β = 0 0 β α() β ( ) = αβ = α β = β = = 0 0β 0 0 0 0 0 0 0 β () α( ) = α() α( ) = β () β ( ) = 0 0 0 0 ( ) = { + } S αβ αβ βα Total 8
Linea Combinations of spin states will be eigenfunctions The two basis vectos αβ and βα ae not eigenfunctions of S-squaed, but the sum and diffeence ae, and have diffeent eigenvalues: ( ± ) = ( ± )( ± ) S αβ βα αβ βα Total The sum then acts like the thid eigenfunction of S-squaed with eigenvalue of = s(s+), so s=, and the diffeence is an eigenfunction of S-squaed with eigenvalue 0, so s=0. Because the spin s= has 3 states (coesponding to ms=,0,-) it is called the tiplet state. The spin s=0 state has only one state so it is a singlet. Without looking at the details moe closely, when combining electons in configuations we can make the esulting states eigenstates of the spin and also of the obital angula momentum opeatos (following the same ecipe). 9
Tem Levels and States Fom a given configuation we can detemine the total z component of the spin and the total z componenet of the obital angula momentum, by simply adding the values of the individual electons in the full electon configuation. This implies making athe abitay choices fo the spins and the m values, but that is OK because we want to descibe all possible aangements of electons. (See Figue 0.0 fo examples). We have discussed how complicated getting the spin ight fo just two electons can be. Same is tue fo the obital states and quantum numbes. A Tem is a goup of states that have the same L and S (total) values, that come fom the same L-squaed and S-squaed values. 0
The Two spin example (again) Fom the configuations fo the ss state (of He) we can deduce the spin states and make the coect deteminants and spin states. m s = 0-0 Fom this deduce that s= coves 3 of these configuations, and the s=0 is the one left ove. So we fom the antisymmetic poducts given the spin states we made befoe (as eigenfunctions of the S-squaed spin opeato) If the spin pat is anti-symmetic the spatial pat must be symmetic (and vise vesa). Pactice witing the thee symmetic states of the spin tiplet and combine it with the pope linea combination of the spatial wave function.
Tem Symbols fo the Obital Angula Momentum Now conside a configuation with moe than electon in a valence obital unapied. Fom the configuation the (nl) numbes of each electon one knows the l value as well as the m_l value. Fom the l values fo say two electons, we can put togethe the vaious possible total L value. The method of addition of the l values ( l and l) is called the Tiangle Rule which follows fom the Clebsch-Godon Coefficients L = l + l, l + l, l l Then this L is used in anothe tiangle ule with etc. The set of L values then ae witten with the capital symbols, S, P, D, F, G etc (countepat to the single electon states). The multiplicity of an L state is L+, and the multiplicity of the spin is S+. S + The tem symbol is witten: L The degeneacy of the tem symbol is (S+)*(L+). I am skipping the discussion on Spin Obit Coupling and the J quantum numbes l 3
A Tem Symbol Example The example (Fig 0.0) of two electons in p states, such as C: s s p Need only conside the electons in the p obitals. L can be 0, o (S,P od) only because of Tiangle Rule How many distinct configuations (o states) ae thee? Fist Electon is 3 time (3p and spin) Second one is one less because of occupancy of fist Total of 30 states but two electons ae indistinguishalbe so thee ae 5 Now find 5 states associated by tems Maximal one is fo two electons in m=, so spins ae paied This is a D state, and s=0, so a singlet D, and it has *5=5 states Next is a P state, one electon in m=, m=0 and unpaied spins This one must be a tiplet P so it has 3 times 3 o 9 states Finally thee is an S state which must have m=0, m=0 so the spins ae paied. So it is a singlet S and only one state. So found all 5 states. Best to wite them out. 3
As a deteminant: This guaantees antisymmetic wave functions φ( ) α( ) φ( ) β ( ) φn ( ) β ( ) φ( ) α( ) φ( ) β ( ) φn ( ) β ( ) (, N ) Ψ = N! ( N) ( N) ( N) ( N) ( N) ( N) φ α φ β φ β Conside electons in the gound state. The configuation is N s s ( ) = φ( ) = φ( ) s() α() s() β () (, ) s( ) α( ) s( ) β ( ) s( ) α( ) s( ) β ( )! s( ) α( ) s( ) β ( ) = s() s( ) { α() β ( ) β () α( ) } Ψ ( ) = Ψ( ) { } Ψ = = Veify that:,,, and show that the wave function is nomalized The Configuation: The values of n and l fo each electon. Fo now we don t list the spin o m values as they don t affect the enegy in this simple pictue. Late we will include them because they do affect the enegy in highe level 4 teatments.
A thee electon system The Slate Deteminant enfoces the Pauli Exclusion pinciple, which is that each electon must have a set of quantum numbes unique to that electon. Choices ae: nlm,, l, ms whee m s = ± Expand the 3 electon deteminant by cofactos. We abitaily chose to put the thid electon into an alpha spin state. ( ) α( ) ( ) β( ) ( ) α( ) ( ) α( ) ( ) β( ) ( ) α( ) ( ) α( ) ( ) β ( ) ( ) α( ) s s s Ψ (,,3) = s s s 3! s 3 3 s 3 3 s 3 3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) { } s( ) α( ) s( ) s( 3) β ( ) α( 3) s( ) s( 3) α( ) β ( 3) = s( ) β( ) α( ) α( 3) { s( ) s( 3) s( 3 ) s( ) } 3! + s( ) α( ) s( 3) s( ) α( ) β( 3) β ( ) α( 3) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) s β s α s α s α s α s β = s() α() s() β () + s() α() 3! s 3 β 3 s 3 α 3 s 3 α 3 s 3 α 3 s 3 α 3 s 3 β 3 This does not guaantee that { } 5
The Vaiational Method The most poweful theoem fo computing the enegies of molecules and atoms. You can vay paametes of the obitals until the enegy gets lowe. The lowest enegy is the best enegy. You can NEVER get a bette enegy than the tue gound state enegy with any tial (i.e. non-exact) wave function. Poof: Expand any tial wave function in the basis set of exact answes (which we don t know but we pesume exists). Of the set of exact enegies the gound state enegy is the lowest. Φ = c Ψ whee c = Ψ Φ d * Tial m m m m Tial m m ( * ) 0 and Theefoe ( * ) ( * ) * * ETial ( cmcm ) Eo = Eo ( cmcm ) = Eo ( ) E = Φ HˆΦ d = Φ HˆΦ d = c Φ HˆΨ d = c c E * * * * Tial Tial Tial Tial Tial m Tial m m m m m cc E E cc E cc E E m m m o m m m m m o Tial m E o m 6
A simple example of the Vaiational Method Fo a paticle in a box, of length a=, in the gound state n= h h a =, φ = sin ( πx) and E = π = = 0.5 m 8 m m 3 5 7 9 sin y = y y + y y + y N ( ) ( ) { } 3 5 7 9 ( x; α) N( α) ( x x ) α x ( x x ) Φ = + + ( ) 0 = 3.63 3 h h E( 0) = N ( 0) 6 x( x x ) dx 0.33 m = = m 5 m x= 0 With no adjustment to alpha we ae about 6% geate than the exact answe. Anothe tial wave function that is symmetic about the cente of the box (and gives a good enegy) is: ( x) Φ= 30 x E Tial h = 0.7 m 7
Enegy Equations with the Obital Appoximation Combine the obital appoximation with the antisymmetic fom and know that we can vay paametes within the obitals to obtain optimized wave functions. Conside just two electons in a common spatial obital with diffeent spins. This is a closed shell poblem and equies only a single deteminant: The point is to find that we have an equation fo each obital and that each electon expeiences the othe in mean field o aveaged way. We obtain exchange and coulombic inteactions between the two electons. 8
The simplest Possible Analysis of He gound state The simplest enegy is to assume that two electons ae in the same s obital with no coections to that obital () s( ) Ψ= s { α() β ( ) β ( ) α( ) } ( ) * E = Ψ H + H + H Ψ d = Es + J e e J s s s s d d s s s s d d * * * * = () ( ) () ( ) ( () () ) ( ( ) ( ) ) 4πε = o, 4πε o, Fom the vaiational pinciple, we can modify the adial pat of the s obital, by choosing an effective Z value fo it. We can only impove ou estimate of the enegy by vaying the effective nuclea chage. ς ao s (, ς ) = e π ao 3 ς As we lowe zeta to educe the J tem, the Es enegy goes up (because the Z= wave function is the exact one fo the s gound state with a He+ nucleus). 9
An excited state of He In the pevious pictue the two electons wee in the same spatial obital. Now, if they ae in diffeent obitals, as in an excitation fom a s to a s (which is fobidden by the selection ules) we can conside whethe a singlet (ie. Spin paied state) o a tiplet (spins unpaied state) is of lage o smalle enegy. To fom a singlet excited state, we needed to add two deteminants to obtain a function that is spatially symmetic and spin antisymmetic. The thee tiplet states ae all of the fom of a spatial antisymmetic fom and a spin symmetic fom. { α( ) β ( ) β ( ) α( ) } s() s( ) s() s( ) Ψ Singlet = + ( ) E = Ψ H + H + H Ψ d = E + E * Snglet s s * e + s() s( ) s() s( ) s() s( ) s() s( ) dd 4πε + + o, E = E + E + J + K Snglet s s e e J = s s s s dd = 4 4πε * * () ( ) () ( ) () () πεo,, s s s s dd * * ( ) ( ( ) ( ) ) e e K s s s s dd s s s s dd * * * * = () ( ) () ( ) ( () () ) ( ( ) ( ) ) 4πε = o, 4πε o, o 0
He ss state cont d Fo the tiplet state, the spatial pat is anti-symmetic and the spin pat is symmetic. This state can be witten as a single deteminant. The enegy of this state is diffeent only in the exchange integal. Because the K (exchange) integal is positive this state is of lowe enegy than the coesponding singlet state. { α( ) β ( ) + β ( ) α( ) } s() s( ) s() s( ) Ψ Singlet = ( ) E = Ψ H + H + H Ψ d = E + E * Tplet s s * e + s() s( ) s() s( ) s() s( ) s() s( ) dd 4πε o, E = E + E + J K Tplet s s e e J = s s s s dd = 4 4πε * * () ( ) () ( ) () () πεo, * * ( ) ( ( ) ( ) ) * * * * = () ( ) () ( ) ( () () ) ( ( ) ( ) ) 4πε = o, 4πε o,, s s s s dd e e K s s s s dd s s s s dd o
The Hatee-Fock SCF Method Each electon is in an obital. Each electon expeiences all of the othe electons in an aveaged sense due to the coulombic epulsion among electons. Geneally the HF method is used with a single deteminant. This is woefully inadequate fo atoms as they most often have open shelled configuations Molecules usually have closed shell configuations, so the single deteminant HF method is often quite satisfactoy fo appoximate popeties such as enegies, dipole moments, polaizabilities, tansitions dipole moments etc. Thee is anothe method called the density functional theoy o DFT that is excellent at computing molecula stuctue and elated popeties, and uns much faste on computes.