By Robert S. Mulliken
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4 5 , 1 8 3 3 ( 1 9 6 6 ) . 3 7 . W. Meyer, / . Chem. Phys. 5 8 , 1 0 1 7 ( 1 9 7 3 ) . 3 8 . R. Ahlrichs, H . Lischka, V. Straemmler, and W. Kutzelnigg, / . Chem. Phys. 6 2 , 1 2 2 5 (1975). 3 9 . (a) W. Meyer, / . Chem. Phys. 64, 2 9 0 1 ( 1 9 7 6 ) . (b) C. E . Dykstra, H. F. Schaefer, III, and W . Meyer, ibid. 6 5 , 2 7 4 0 ( 1 9 7 6 ) . (c) C. E. Dykstra, H . F . Schaefer, III, and W. Meyer, ibid. 65, 5 1 4 1 ( 1 9 7 6 ) . CHAPTER II ONE-ELECTRON M O s AS P R O T O T Y P E S A. H + 2 M O s IN ELLIPTICAL COORDINATES + E x p e r i m e n t a l l y , t h e H 2 i o n is well k n o w n in m a s s s p e c t r o m e t r y , b u t t h e o n l y o t h e r e x p e r i m e n t a l d a t a a r e o n h y p e r f i n e s t r u c t u r e in its v i b r a t i o n r o t a t i o n levels [ 1 ] .
EXACT CALCULATIONS O N H 2 H 2 is t h e s i m p l e s t p r o t o t y p e o f p o l y e l e c t r o n h o m o p o l a r d i a t o m i c m o l ecules. H e r e , ^ = - ± V - ± V 2 2 r ai - - - r- - - -r bl a2 b2 +r - , r (1) l2 + differing f r o m of H 2 because there a r e t w o electrons, which bring in the n e w type of complication involved in the interelectronic repulsion term l / r 1 2. S o l u t i o n s o f t h e S c h r o d i n g e r e q u a t i o n h a v e b e e n o b t a i n e d f o r s e v e r a l o f t h e l o w e s t e n e r g y s t a t e s i n a f o r m first u s e d b y J a m e s a n d C o o l i d g e [ 1 ] , a n e x p a n s i o n i n elliptical c o o r d i n a t e s t o g e t h e r w i t h r 1 2, a s Σ A i + ^ ] ) ( A i % V ^ V + V ^ i V p a (2) mnjkp + w h e r e λ a n d μ a r e a s defined f o r H 2 i n C h a p t e r I I , t h e s u b s c r i p t s refer t o t h e t w o e l e c t r o n s , a n d ρ = 2rl2/R.
POPULATION 37 ANALYSIS w h e r e χα a n d xb m a y b e Μ A O s . N o w if E q . (12) is v a l i d , 2 Na = ( c o s γ + S sin γ c o s y)/(l + S sin 2y) = ± [ 1 + c o s 2 y / ( l + 5 sin 2 y ) ] . (14) A n e x a m i n a t i o n o f t h e c o n s e q u e n c e s o f E q . (14) s h o w s t h a t i n c e r t a i n r a n g e s o f y a n d S, Na is n e g a t i v e a n d , since Na + Nb= 1 for H e H 2 + , Nb is t h e n g r e a t e r t h a n 1. B o t h t h e s e r e s u l t s a r e o b v i o u s l y n o n s e n s e for t h e g r o s s p o p u l a t i o n s o n a o r b.