lecture27 - 5.61 Physical Chemistry Lecture#27 page 1 MANY...

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5.61 Physical Chemistry Lecture #27 page 1 MANY ELECTRON ATOMS Thus far, we have learned that the independent particle model (IPM) gives a qualitatively correct picture of the eigenstates of the helium atom. What about atoms with more than two electrons, such as lithium or carbon? As it turns out, the IPM is capable of giving a realistic picture of atomic structure in essentially an analogous fashion to the helium case. To begin with, we set up our coordinates so that the nucleus is at the origin and the N electrons are at positions r 1 , r 2 , r 3 , … r N . In terms of these variables, we can quickly write down the many-electron Hamiltonian (in atomic units): ˆ 1 N 2 N Z N N 1 H = j + ∑∑ 2 j = 1 j = 1 r ˆ j i = 1 j > i r ˆ i r ˆ j Kinetic Energy Electron-Nuclear Electron-Electron Attraction Repulsion Thus, the Hamiltonian has the same three sources of energy as in the two electron case, but the sheer number of electrons makes the algebra more complicated. As before, we note that we can make the Hamiltonian separable if we neglect the electron-electron repulsion: H ˆ = − 1 N 2 N Z = N 1 2 Z N h ˆ NI 2 j r ˆ 2 j r ˆ i j = 1 j = 1 j j = 1 j j = 1 where each of the independent Hamiltonians h ˆ i describes a single electron in the field of a nucleus of charge +Z . Based on our experience with separable Hamiltonians, we can immediately write down the eigenstates of this Hamiltonian as products with energies given as sums of the independent electron energies: Ψ = ψ 1 ψ 2 ψ 3 ... ψ N E = E + E + E + ... + E Where (1) is a shorthand for ( r 1 , σ 1 ) and k { n , l , m , s } specifies all the quantum k 1 ( ) k 2 ( ) k 3 ( ) k N ( ) k 1 k 2 k 3 k N i i i i i numbers for a given hydrogen atom eigenstate. Of course, there is a problem with these eigenstates: they are not antisymmetric. For the Helium atom, we fixed this by making an explicitly antisymmetric combination of two degenerate product states: ψ ( 1 ) ψ ( 2 ) 1 1 ψ 1 ψ 2 ψ 2 ψ 1 = ( 1 s α ( ) 1 s β ( ) 1 s α ( ) 1 s β ( ) ) ψ 1 1 s s α β ( ) 1 ψ 1 1 s s α β ( ) 2 2 2
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5.61 Physical Chemistry Lecture #27 page 2 where on the right we have noted that this antisymmetric product can also be written as a determinant of a 2x2 matrix. As it turns out, it is straightforward to extend this idea to generate an N particle antisymetric state by computing an NxN determinant called a Slater Determinant : ψ ( 1 ) ψ ( 1 ) ψ ( 1 ) ψ ( 1 ) k k k k 1 2 3 N k 1 ( ) k 2 ( ) k 3 ( ) k N ( ) ψ 2 ψ 2 ψ 2 ψ 2 Ψ ( 1,2,..., N ) = ψ 3 ψ 3 ψ 3 ψ 3 k ( ) k ( ) k ( ) k ( ) 1 2 3 N � � � k 1 ( N ) ψ k 2 ( N ) ψ k 3 ( N ) ψ k N ( N ) 1 ! N ψ As you can imagine, the algebra required to compute integrals involving Slater determinants is extremely difficult. It is therefore most important that you realize several things about these states so that you can avoid unnecessary algebra: A Slater determinant corresponds to a single stick diagram. This is easy to see by example: 2p x 1 s α ( 1 ) 1 s β ( 1 ) 2 s α ( 1 ) 2 p α ( 1 ) x s α ( ) 1 β ( ) 2 s α 2 2 p α ( ) 1 2 s 2 ( ) 2 x 2s Ψ ( 1, 2,3, 4 ) = 1 s α 3 1 s β 3 2 s α
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