and Slater Sum Rule Method * M L = 0, M S = 0 block: L L+ L 2

Similar documents
Objectives. We will also get to know about the wavefunction and its use in developing the concept of the structure of atoms.

Lecture 1. time, say t=0, to find the wavefunction at any subsequent time t. This can be carried out by

Three-dimensional systems with spherical symmetry

Jackson 4.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

The Solutions of the Classical Relativistic Two-Body Equation

Problem Set 10 Solutions

Mechanics Physics 151

Mechanics Physics 151

Angular Momentum Theory.

5.61 Physical Chemistry Lecture #23 page 1 MANY ELECTRON ATOMS

Lecture 7: Angular Momentum, Hydrogen Atom

Physics 2D Lecture Slides Dec 1. Vivek Sharma UCSD Physics

Jackson 3.3 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Physics 505 Homework No. 9 Solutions S9-1

Pressure in the Average-Atom Model

Seidel s Trapezoidal Partitioning Algorithm

Many Electron Atoms. Electrons can be put into approximate orbitals and the properties of the many electron systems can be catalogued

Doublet structure of Alkali spectra:

FI 2201 Electromagnetism

3.23 Electrical, Optical, and Magnetic Properties of Materials

= ρ. Since this equation is applied to an arbitrary point in space, we can use it to determine the charge density once we know the field.

Mechanics Physics 151

3.23 Electrical, Optical, and Magnetic Properties of Materials

J. N. R E DDY ENERGY PRINCIPLES AND VARIATIONAL METHODS APPLIED MECHANICS

PHYS 705: Classical Mechanics. Central Force Problems I

Appendix A. Appendices. A.1 ɛ ijk and cross products. Vector Operations: δ ij and ɛ ijk

3. Electromagnetic Waves II

PHYS 705: Classical Mechanics. Central Force Problems II

Stanford University CS259Q: Quantum Computing Handout 8 Luca Trevisan October 18, 2012

Quantum theory of angular momentum

3.012 Fund of Mat Sci: Bonding Lecture 5/6. Comic strip removed for copyright reasons.

Advanced Quantum Mechanics

Nuclear models: Shell model

Nuclear and Particle Physics - Lecture 20 The shell model

11) A thin, uniform rod of mass M is supported by two vertical strings, as shown below.

ON THE TWO-BODY PROBLEM IN QUANTUM MECHANICS

PROBLEM SET #1 SOLUTIONS by Robert A. DiStasio Jr.

Many Electron Theory: Particles and holes. Unitary transformations.

c n ψ n (r)e ient/ h (2) where E n = 1 mc 2 α 2 Z 2 ψ(r) = c n ψ n (r) = c n = ψn(r)ψ(r)d 3 x e 2r/a0 1 πa e 3r/a0 r 2 dr c 1 2 = 2 9 /3 6 = 0.

1 Spherical multipole moments

3.320: Lecture 6 (Feb ) Photos of Hartree, Fock, Hohenberg, Kohn, and Sham removed for copyright reasons.

Q. Obtain the Hamiltonian for a one electron atom in the presence of an external magnetic field.

Physics 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel. Problem Set 7 Maximal score: 25 Points. 1. Jackson, Problem Points.

Question 1: The dipole

Vector d is a linear vector function of vector d when the following relationships hold:

Merging to ordered sequences. Efficient (Parallel) Sorting. Merging (cont.)

Matrix Elements of Many-Electron Wavefunctions. noninteger principal quantum number. solutions to Schröd. Eq. outside sphere of radius r

3D-Central Force Problems I

5.111 Lecture Summary #6 Monday, September 15, 2014

EQUATIONS OF MOTION LUCA GUIDO MOLINARI

( n x ( ) Last Time Exam 3 results. Question. 3-D particle in box: summary. Modified Bohr model. 3-D Hydrogen atom. r n. = n 2 a o

As is natural, our Aerospace Structures will be described in a Euclidean three-dimensional space R 3.

Physics 161 Fall 2011 Extra Credit 2 Investigating Black Holes - Solutions The Following is Worth 50 Points!!!

= e2. = 2e2. = 3e2. V = Ze2. where Z is the atomic numnber. Thus, we take as the Hamiltonian for a hydrogenic. H = p2 r. (19.4)

Relating Scattering Amplitudes to Bound States

6 PROBABILITY GENERATING FUNCTIONS

3D-Central Force Problems II

MATH 220: SECOND ORDER CONSTANT COEFFICIENT PDE. We consider second order constant coefficient scalar linear PDEs on R n. These have the form

Problem set 6. Solution. The problem of firm 3 is. The FOC is: 2 =0. The reaction function of firm 3 is: = 2

Vector Spherical Harmonics and Spherical Waves

A New Approach to General Relativity

SIO 229 Gravity and Geomagnetism. Lecture 6. J 2 for Earth. J 2 in the solar system. A first look at the geoid.

A Relativistic Electron in a Coulomb Potential

Why Professor Richard Feynman was upset solving the Laplace equation for spherical waves? Anzor A. Khelashvili a)

sinγ(h y > ) exp(iωt iqx)dωdq

Fields and Waves I Spring 2005 Homework 4. Due 8 March 2005

( ) ( ) Last Time. 3-D particle in box: summary. Modified Bohr model. 3-dimensional Hydrogen atom. Orbital magnetic dipole moment

1D2G - Numerical solution of the neutron diffusion equation

( ) into above PDE. ( ), wherec = 1

New problems in universal algebraic geometry illustrated by boolean equations

Chemical Engineering 412

Topics in Nuclear Astrophysics

Calculation of Quark-antiquark Potential Coefficient and Charge Radius of Light Mesons

This gives rise to the separable equation dr/r = 2 cot θ dθ which may be integrated to yield r(θ) = R sin 2 θ (3)

A Hartree-Fock Example Using Helium

Temporal-Difference Learning

APPENDIX. For the 2 lectures of Claude Cohen-Tannoudji on Atom-Atom Interactions in Ultracold Quantum Gases

Geometry of the homogeneous and isotropic spaces

ASTR415: Problem Set #6

Probability Estimation with Maximum Entropy Principle

Homework # 3 Solution Key

Quantum Chemistry Notes:

Preliminary Exam: Quantum Physics 1/14/2011, 9:00-3:00

1.2 Differential cross section

3.012 Fund of Mat Sci: Bonding Lecture 11. The future of electronics? A pentacene molecule deposited on SiO 2. Image removed for copyright reasons.

Computational Methods of Solid Mechanics. Project report

I. CONSTRUCTION OF THE GREEN S FUNCTION

Hopefully Helpful Hints for Gauss s Law

Rydberg-Rydberg Interactions

Physics 235 Chapter 5. Chapter 5 Gravitation

15.081J/6.251J Introduction to Mathematical Programming. Lecture 6: The Simplex Method II

L6 Energy Conversion. EIEN20 Design of Electrical Machines, IEA, Previous lecture. L6: Energy conversion. Today s goal

b Ψ Ψ Principles of Organic Chemistry lecture 22, page 1

Unit 6 Lecture Number 36

( ) [ ] [ ] [ ] δf φ = F φ+δφ F. xdx.

Anyone who can contemplate quantum mechanics without getting dizzy hasn t understood it. --Niels Bohr. Lecture 17, p 1

Lecture 8 - Gauss s Law

Solution to HW 3, Ma 1a Fall 2016

Chem 453/544 Fall /08/03. Exam #1 Solutions

Let us next consider how one can calculate one- electron elements of the long- range nuclear- electron potential:

Transcription:

5.7 Lectue #4 e / ij and Sate Sum Rue Method 4 - LAST TIME:. L,S method fo setting up NLM L SM S many-eecton basis states in tems of inea combination of Sate deteminants * M L = 0, M S = 0 boc: L L+ L S S S + * diagonaize S (singets and tipets) * diagonaize L in same basis that diagonaizes S [Reca: to get matix eements of L, fist evauate L and then eft mutipy by ψ j. couped epesentations njωs and NJLSM J. Pojection opeatos: automatic pojection of L eigenfunctions * emove unwanted L pat * peseve nomaization of wanted L pat * emove oveap facto ψ i TODAY:. Sate Sum Rue Tic (tace invaiance): MAIN IDEA OF LECTURE.. evauate e ij matix eements (tedious, but good fo you) i> j [-e opeato, spatia coodinates ony, scaa wt J,L,S] * mutipoe expansion of chage distibution due to othe eectons * matix eement seection ues fo e / ij in both Sate deteminanta and many-e basis sets * Gaunt Coefficients (c ) (tabuated) and Sate-Condon (F,G ) Couomb and Exchange paametes. Because of sum ue, can evauate mosty ab ab and ab ba type matix eements and neve ab cd ij ij ij type matix eements.. Appy Sum Rue Method 4. Hund s st and nd Rues updated Septembe 9,

5.7 Lectue #4 4 -. Sate s Sum Rue Method It is amost aways possibe to evauate e / ij matix eements without soving fo a LM L SM S basis states. * tace of any Hemitian matix, expessed in ANY epesentation, is the sum of the eigenvaues of that matix (thus invaiant to unitay tansfomation) * e / ij and evey scaa opeato with espect to J ˆ (o L,S) ˆ ˆ has i> j nonzeo matix eements diagona in J and M J (o L and M L ) and independent of M J o (M L,M S ) [W-E Theoem: J is a GENERIC ANGULAR MOMENTUM with espect to which e / ij is cassified] Reca fom definition of, that e / ij is a scaa opeato with espect to ˆ ˆ ˆ J, L, S but not with espect to j i o i. Inteeectonic Repusion: e ij i> j * destoys obita appoximation $$ fo eectonic stuctue cacuations * coeation enegy, shieding e e e e at (, θ, φ ) at (, θ, φ ) scaa with espect to J, L, S, s i but not j i, i = = + cos, [ ] / = + ( ) updated Septembe 9,

5.7 Lectue #4 expand as powe seies in < > whee < is smae of, ( integas evauated in egions : <, < ) 4 - engthy ageba see Eying, Wate, and Kimba Quantum Chemisty pages 69-7 and, fo eationship between Legende poynomias and Y m ( θφ, ), pages 5-59. mutipoe expansion n 4 = π n+ ij n= 0 m=n n < m m Y Y n n i i n j j (, ) (, + ) > [ ] θ φ θ φ not pincipa q.n.! convegent seies n -poe moment (n=0 monopoe, n= dipoe, ) angua momenta magnitude n, pojection m scaa poduct of angua momenta, one fo i-th patice, one fo j-th * * convets m to m n-poe chage distibution n-th an tenso n+ components No dependence on s, so / ij is scaa with espect to S, si, sj. m [ Yn ( θi, φi)= θi, φi i = n, m = m i ] updated Septembe 9,

5.7 Lectue #4 4-4 The eason fo this athe compicated ooing expansion is that it is we suited fo integas ove atomic obitas which ae expessed in tems of i, θ i, φ i, which ae coodinates of the i-th e with espect to the cente of symmety (nuceus) athe than the othe e. It enabes use of AO basis states. Othewise / ij integas woud be nightmaes. Seection ues fo matix eements: not pincipa q. n. i n, m = m, m = i s 0 i obitas j n, m = m, m = j s 0 j tiange ue, n + i i i i (steps of because of paity) tem in mutipoe expansion ovea: L = 0, S = 0, M L = 0, M S = 0, and indep. of M L, M S Can use any M L, M S fom box diagam. It is aso cea how to evauate the angua factos of the atomic obita matix eements using -j coefficients. Specia tabes of Gaunt Coefficients (aso C&S pages 78-79, Goding, page 4, see handout). updated Septembe 9,

5.7 Lectue #4 4-5 genea / matix eement ( so = 0,, and ae possibe) e ± ab cd = ab cd ab dc e ± ab cd = m m m m m m m m (, ) (, ) +, + { = 0 tenso an fo poduct of AOs occupied by e ± # must be same as fo # fo scaa poduct of n-th an tensos δ δ δ ( ) s s s s a c b d a b c d / does not opeate ** ** on spin coodinates GAUNT COEFFICIENTS ANGULAR FACTOR OF INTEGRAL ( ) R n n n n a a b b c c d d adia facto / scaa with espect to ˆ = ˆ + ˆ L ( a a c ) ( c b b d ) d e e c m, m c m, m / c ( m, m ) + A A 000 + tabuated m m, m, m Cebsch-Godan coefficients that esut fom intega ove poduct of spheica hamonics one fom opeato, two fom obitas tiange ue:- + + + = even ( fom popeties of A 000 ) ( paity) updated Septembe 9,

5.7 Lectue #4 estictions on and m: e intega m + m = m m n m Y n m tiange ue 4-6 fo intaconfiguation matix eements, R (abcd) taes on especiay simpe fom (because the same one o two obitas appea in the ba and in the et). ( ) ( ) ( ) ( ) R ab, ab F a, b R abba, G ab, Sate Condon paametes (these ae educed matix eements dependent ony on a, b, c, d and not on any of the m quantum numbes.) A L-S states fom one configuation ae expessed in tems of the same set of F, G paametes. ab e ab J ab, δ m, m K ab, = ( ) ( ) ( ) DIRECT s a spins must match o K tem vanishes s b EXCHANGE ( ) = ( ) ( ) ( ) J ab ab e ab c m m c m m F,,, n, n, ( ) = δ( sa s ) b [ ( a a b ) b ] ( a a b b) = 0 b ( am, bm a b ) K ab ab e ba m m c m m G,,, n, n fo specia cases, such as nd, n = n and F = G Now we ae eady to use tabes of c (o, moe convenienty, a and b to set up e = 0 ij matix a a a a b b a m, m a a b b b ( a a b ) b b a a b b [ a*( ) a( ) Opb ˆ *( ) b( ) d τ dτ] chage distibutions [ a*( ) b( ) Opa ˆ ( ) b*( ) d τ dτ] something ese! ) updated Septembe 9,

5.7 Lectue #4 4-7 Easy exampe: nf ( eca I, H, G, F, D, P, S) I 60 = αβ H 5 = αα I e = ( ) ( ) ( ) ( ) ( ) = 046,,, = 046,,, [ c, ] F ( nf, nf ) [ ] ( ) I c, c, F nf, nf δαβ, c, G nf, nf = ( ) these ae the ony L-S states epesented by a singe Sate deteminant extemes of M L,M S box diagam since e / ij is a scaa opeato with espect to Lˆ, Sˆ, Jˆ, matix eements ae M L, M S, M J independent so we can use any M L,M S component to evauate the matix eement whicheve is most convenient! H e = 046,,, e m e = 0 one spin α othe spin β {[,, ] (, ) [ (, ) ] (, )} both spins α e e H c c F nf nf c G nf nf = ( ) ( ) (a,a) (b,b) (a,b) Hee is whee eveyone maes mistaes! F ( nf ) F (nf ) Use tabe of c in Goding/C&S handout (C&S page 79). Note that [/76 64] / is impicit afte the fist enty fo f, = 6. convenient facto = 0 4 6 c (,) / / [/76 64] / c (,) 0 7/ [6/76 64] / c (,) 0 +/ 0 / / [7/76 64] / D 5 089 = 76 64 updated Septembe 9,

5.7 Lectue #4 4-8 D is a facto that simpifies the expessions. Each tem has the fom F /D. Ca this atio F. Get simpe ooing expessions when you epace F by D F (D appeas in denominatos of c as [ /D ] / ) I e I = F + F F F + 9 + 76 64 0 4 6 = F + 5F + 9F + F 0 4 6 Aways have two factos of c. Thus F gets divided by D to yied F. H e H = F + ( ) ( ) F + F F 0 4 6 0 7 0 6 7 + 76 64 5 = 9 ( ) 76 64 = F 5F 5 F F 0 4 6 F F F F 0 4 6 A ot of boo eeping, but easy to ean how to use tabes of c, a, b, D. But it is much moe wo fo f than fo f. SUM RULE METHOD: Basic idea is that the sum of diagona eements in the singe Sate deteminant basis set within an M L, M S box is equa to the sum of the eigenvaues! Loo at ML =, MS = box: α0 α and αα. This box geneates H and F, but tace is E( H) + E( F) and we aeady now E( H)! So E E I H E F α0α αα E H E G αβ βα αβ E I E H E ( ) = ( ) = ( ) = + + + D αβ αα ( ) = + ( ) ( )= + + ( ) ( ) α β β α α0β β0α αβ E I E G) ( ) E H E( F) + ( ) ( ( ) = α α + α α + α0α ( ) ( ) ( )= S+ ( ) E P E H E F E S sum of seven sum of six E L updated Septembe 9,

5.7 Lectue #4 4-9 This seems athe aboious, but it is much easie than: * geneating each LM L = L SM S = S as an expicit inea combination of Sate deteminants * then cacuating matix eements of e / ij, because thee ae many nonzeo offdiagona matix eements between Sate deteminants in the same M L,M S box. Hee is the fina esut fo the enegies of a (nf) S+ L tems: 0 E = E + E + E ( 0) ( 0) Z R E = sum of obita enegies fom h = = n ( ) E = e SO + H ij next ectue ( ) E intaconfiguationa spin - obit inteconfiguationa e = ( )+ ( ij) Fo nf ( ) ( ) ( ) eady now shieded by a fied subshes shieded by same subshe ε CI n Bae nuceus hydogenic obita enegy o paty shieded by fied shes. I ε nf 0( ) F nf F nf ( ) F( nf ) F( nf ) + + 5 + 9 + 4 6 H nf G nf F nf D nf P nf S nf ε + F 0 5 F 5 F 4 F 6 ε + F 0 0 F + 97 F 4 + 78 F 6 ε + F 0 0 F F 4 86 F 6 ε + F 0 + 9 F 99 F 4 + 75 F 6 ε + F 0 + 45 F + F 4 87 F 6 ε + F 0 + 60 F + 98 F 4 + 76 F 6 shieded-coe configuationa enegy intaconfiguation L-S tem spittings (thee is NO cente of Gavity Rue fo degeneacy weighted L-S tems) updated Septembe 9,

5.7 Lectue #4 4-0 Now it is easy to show that a F s ae > 0 and F >> F + etc. (oughy facto of 0 pe step in ) Fom this we get an empiica ue Lowest E of a L S tems is the one with * MAXIMUM S * of those with Maximum S, owest is the one with MAXIMUM L These ae Hund s fist and second (of thee) ues. Note aso that Hund s ues do nothing about pedicting the enegy ode of L-S tems except fo the identity of the singe, owest enegy L-S tem. updated Septembe 9,

5.7 Lectue #4 Nonectue 4 - Thee ae sevea inteesting pobems aso soved by this e / ij fomaism.. Enegy spittings between and Sate deteminanta chaactes of two o moe L,S tems of the same L and S that beong to the same L,S configuation e.g. d two D tems see pages 47-50 of Goding fo secua deteminant fo D of d. matix eements of e / ij between same L,S tems that beong to two diffeent configuations e.g. nd ndn d S, P, D, F, G S, P, D, F, G no Paui estictions SPDFG,,,, so thee wi be S~ S P~ P D~ D F~ F G~ G inteconfiguationa CI s, and each of these 5 inteaction matix eements wi NOT be of the same magnitude. updated Septembe 9,