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X;) S ft(x,)*g(x„ . . , x,) = 0 (3-36) Since the F form an orthogonal set, the coefficient of each must vanish, leaving i S (fiix^gix^ x l . . , x) = 0 b C. /x , . · · , x„) S φ,(χ,)*χ,(χ,) = 0 2 (3-38) Consequently, if F φ 0 and Gj φ 0, then i S φ,(χ,)*χ/χ,) = 0 x l (3-39) J Since by (3-33) the occupation number of φ, in / i s | | ^ | | , it follows that all the occupied natural orbitals of / are orthogonal to all the occupied natural orbitals of g. That is, the two sets of natural orbitals lie in disjoint orthogonal subspaces.

V l + V 2 * ' ( 2 , . . , # ) and i> the number of M 2 A. ,7V). Then + , 5, M _ Μ J / s- \ M+ ι γ S Ï,M U + ) 25 + 2 J> T < s + "-"' Ρ Λ V + M+i I < + ^ y 25 + 2 ) (4-33) 2 With this choice of the χ ; · , Tr |x * " > < x * " ' | ν < v M = AT v ? )]• 25 \β)(α [(S - M)(S + M + 1)] PJ, M' = M + 1 - M + 1)]* 25 ' Λ/' = M + 1 [(5 + M ) ( 5 - Λ/ + 1)]· M' = M - 1 25 + 2 Μ' = M - 1 I«XJS| 0, < 25 + 2 [(5 + M)(S \«)<β\ ρ Ρ > ^p otherwise (4-34) S ° ΡΙΛ/Λ/' = 0 f ° 1^ ~ ^'1 ^ 2. ]( ^ - (4-35) ) or PIMM ( S + M \ l-l^J ~ ( , / S - A/ + 1 \ 1 P s -i l + S —M \ 2S 25 + 2 ) * P α)(α| + .

Since V involves a two-body operation, (4-96) is a boundary condition involving p and p . It is probably more customary to pick X to be one component of the Cartesian coordinates describing the position of nucleus a and to pick the r, to be measured from a fixed point in space so that for an Η of the form (4-70) { 2 a \ n = - 2<7„(r, - R > , ; + q 3 a i Σ β ¥ « q {* P P - RJ V (4-ioo) 46 Analytic Properties of the First-Order Reduced Density Matrix 4 Consequently, the classical electric field at the position of nucleus a is given by δ = - α Σ f + V«" P « 3 T dr (4-101) Equation (4-101) appears to give the electric field at the nucleus by a classical formula involving the fields generated by the other nuclei and the field produced by TV units of charge distributed continuously over space with density p (r).

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