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From the course PHYS261, University of Bergen. This is a PDF with links. The links do not work here on slideshare preview.

### Transcript of Many electrons atoms_2012.12.04 (PDF with links

• 1. Many electron atoms ( x ) x3 dV = d xxrrAutumn 2012 Version: 04.12.2012 List of topics

2. (1) Pauli-principle Hunds rule Antisymmetric functions for n-particles - Slater Determinants Filling Shells Figure Ionization energies - Figure Hartree - Selfconsistent eld - Iterations only Iterations part Selfconsistent eld Result - Figures Congurations - Ionization potentials of atoms - Tables Screened potential and Centrifugal barrier - Explains Tables Energy for N-particles - Using Slater Determinants Helium example Lithium example Counting nonzero terms N particles - Energy Summary Schrdinger equation from variational method o Variational method - deriving Hartree-Fock Equations Hartree-Fock Equations Total energy and the selfconsistent orbital energies Evaluation of electron repulsion Conguration mixing21 3 2 2 8 4 3 3 2 5 1 2 1 1 2 1 3 2slide + 3 slides slides slides slides slides slides slides tables slides slides slide slides slide slide slides slide slides slides + 2slides 3. 1Filling up the shells with electronsIn order to know and to, more or less, understand in which state an electron is bound we can use some basic rules. The generel idea is that the lowest energy-state is the most stable one. The exited states can in most cases fall in the lower state or ground state by emission of light. The principles one need to know how to ll up the shells are the Pauli-principle and the Hunds rule. List of topics3 4. 2Pauli-principleThe lling of the shells observed was explained by Pauli by formulating the Pauli exclusion principle: Two electrons can not be in the same state - dened here by possible quantum numbers n, l, m and ms n, the principal quantum number of the shell, l, the angular momentum quantum number of the subshell, m, the magnetic quantum number ms denotes spin up and spin down, ms = 1/2 or ms = 1/2 The Pauli principle mathematical: require the multiparticle wavefunction antisymmetric with respect to exchange of two particles. If two particles coordinates are exchanged, the function must change sign. If two particles are in the same state (function), this requirement leads to zero wavefunction, thus impossible. List of topics4 5. Figure 1: Atomic levels (shells)List of topics5 6. Figure 2: Atomic levels (shells) lled up to the element ....List of topics6 7. 3Antisymmetric product function for n-particles (1)P (perm(,,...)) perm ( .... ) (x1 )(x2 )....(xn )(x1 , x2 , ...xn ) = perm(,,...)where each term in the sum looks as (x1 )... (x2 )... ..., summing over all permutations, and P (perm(, , ...)) is the number of swaps of the given permutation perm(, , ...) This is very close to the denition of the determinant ndet(A) =sgn()Ai,(i) i=1SnThe above in this notation n(x1 , x2 , ...xn ) =sgn() SnList of topics7(i) (xi ) i=1 8. Slater determinant The antisymmetric combination for n-particles can be written as a determinant in this way: (x1 ) (x2 ) ... (xn ) (x1 ) (x2 ) ... (xn ) ... ... ... ... (x1 ) (x2 ) ... (xn ) For 3 particles: 3 particle Slater determinant (x1 ) (x2 ) (x3 ) (x1 ) (x2 ) (x3 ) (x1 ) (x2 ) (x3 ) List of topics8 9. 3 3 determinant (1) (2) (3) (1) (2) (3) (1) (2) (3) (1) (2) (3) (1) (2) (3) (1) (2) (3) (1) (2) (3) (1) (2) (3) (1) (2) (3) + (1) (2) (3) (1) (2) (3) + (1) (2) (3)List of topics9 (1) (2) (3) (1) (2) (3) 10. 4Hunds ruleHunds rule is the manifestation of the same eect as we have seen in the parahelium - orthohelium eect ( the second rule ) Hunds rst rule Full shells and subshells do have a total circular momentum of zero. This can be calculated and is allways valid. Hunds second rule The total spin S should allways have the highest possible value. So as many of the single electron spins as possible should be parallel. The second rule appears more empirical and applies to a dierent magnetic quantum number of the electrons with parallel spins. If is of course not allowed to break the Pauli-principle. List of topics10 11. 5Example for the Hunds ruleFigure 3: Electron spins of the rst elements. The elements carbon C, nitrogen N and oxygen O are those where the Hunds rule have the biggest inuence. We see electrons with parallel spins in the states m = 1, 0, +1, depending on the element. The magnetic quantum number m gives more or less the direction of the circular moment l. The s-states are the states where l = 0 and the p-states those where l = 1. The names for l = (2, 3, 4, ...) are (d, e, f, ...). The shell-name of the shell n = (1, 2, 3, 4, ...) is (K, L, M, N, ...). List of topics11 12. In cases with more electrons do we get some exceptions. For example is the 4s subshell earlier lled than the 3d. This is caused by the smaller distance of the 4s to the core and thus by the lower energy, which is more stable. The mentioned rules work well for general considerations and for atoms with not to many electrons.6Number of statesUsefull to know is the largest possible number of states for a given n which means until a special shell is lled. n1(2l + 1) = 2n2 .Nmax = 2 (2)l=0We get this formula by adding all the possible quantum number congurations: We get for each n every l in the range (0, 1, ..., n 1), for each l every m in the range (l, ..., 1, 0, 1, ..., l) and for each of these states 2 spin possibilities. List of topics12 13. 7Ionization energiesFigure 4: Ionization energies. The shell properties would explain the structure in general - Periodic table and the Selfconsistent eld List of topics13 14. Especially at the rst elements, we see the minimums at the elements with just one electron in the last shell and a maximum at the noble gas elements. The weak maxima are caused by lled subshells or by the maximum of parallel standing spins. After argon, it is more dicult to observe general tendences. The shell properties would explain the structure in general But there should be no closed shell at argon Details - Periodic table and the Selfconsistent eldList of topics14 15. 8Hartree - Selfconsistent eldInteraction energy of two charges depends on their distance|r x|: W (|r x|) =q1 q2 |r x|The two charges are an electron and a little volume dV at x containing charge cloud of density q1 (e) q2 (x)dV (x)d3 x ( x ) xThe interaction energy of these two charges is dW (|r1 r2 |) =3 dV = d xxr(e)(r2 ) 3 d r2 |r x| rList of topics15 16. Interaction with a cloud; summing over all the small volume elements - it means integrating over the whole volume of the cloud gives the potential energy W (r) =(e)(x) 3 d x |r x|( x ) x3 dV = d xxrr List of topics16 17. If the charge cloud represents one electron in state i (x)and again integrating gives the potential energy due to the interaction with a (probability based density) cloud of electrons(x) = (e)|i (x)|2 If we have N electrons, each in its state, the total density becomes(e)2N|i (x)|(x) = (e)W (r) =2i=1( x ) x3 dV = d xxrrList of topics17N|i (x)|2i=1|r x|d3 x 18. Now solving the Schrdinger equation with W (r), o (T + V + W ) i (x) = Ei i (x) We rst need to know the W (r), but that depends on all the other N solutions N(e)2|i (x)|2i=1W (r) =|r x|d3 xApproximation chain: First we choose some simple approximation, e.g. the hydrogenlike states, or we might know the states foranother atom. We call it (0)i (x) (0)From the set of all N iwe construct e2W (1) (r) =N i=1(0)|i (x)|2|r x|d3 xIn atomic units the whole Schrdinger equation is o 1 22Z (1) (1) (1) + W (1) (r) i (x) = Ei i (x) rList of topics18 19. (0)1. step - choose arbitrary set of iN (0)i (x) 1 22(1)2. step: take the set of iW (1) (r) =i=1|r x|from the 1. step(1)i (x) 1 22d3 xZ (1) (1) (1) + W (1) (r) i (x) = Ei i (x) rN(0)|i (x)|2W (2) (r) =i=1(1)|i (x)|2 |r x|d3 xZ (2) (2) (2) + W (2) (r) i (x) = Ei i (x) rList of topics19(3) 20. (2)3. step: take the set of ifrom the 2. step N(2)i (x) 1 22(3)4. step: take the set of iW (3) (r) =i=1|r x|from the 3. step(3)i (x) 1 22d3 xZ (3) (3) (3) + W (3) (r) i (x) = Ei i (x) rN(2)|i (x)|2W (4) (r) =i=1(3)|i (x)|2 |r x|d3 xZ (4) (4) (4) + W (4) (r) i (x) = Ei i (x) rList of topics20 21. (4)5. step: take the set of ifrom the 4. step N(4)i (x) 1 22(5)6. step: take the set of iW (5) (r) =i=1|r x|from the 5. step(5)i (x) 1 22d3 xZ (5) (5) (5) + W (5) (r) i (x) = Ei i (x) rN(4)|i (x)|2W (6) (r) =i=1(5)|i (x)|2 |r x|d3 xZ (6) (6) (6) + W (6) (r) i (x) = Ei i (x) rList of topics21 22. 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