1056 Y. PAULI DEFORMATION Fig. Y.2. The locality of the Pauli deformation (diagram). (a) Two polymeric chains A and B (with electronic densities in the form of the elongated rectangles corresponding to the isolated molecules A and B) approach one another (b) the Pauli deformation consists of the two density gains (the rec- tangles with +) and a single electron loss (the rectangles with −). Let us assume that the surfaces of the rectangles are equal to the corresponding integrals of the charge distributions −4S/(1 −S 2 )ab in the contact region, 2S 2 /(1 − S 2 )a 2 on molecule A and 2S 2 /(1 − S 2 )b 2 onpolymerB–thisiswhy the electron density loss has a rectangle twice as large as any of the electron density gains (c) a partial Pauli deformation: the density gain 2S 2 /(1−S 2 )a 2 for molecule A has been added to the initial density distribution, and similarly for molecule B (the rectangles became larger, but locally the corresponding increase is small). (d) In order to represent the total Pauli deformation from the result obtained at point c we subtracted the density distribution 4S/(1−S 2 )ab which is located in the contact region. As a result the Pauli deformation, when viewed locally, is large only in the contact region. The only thing that has been changed with respect to the hydrogen molecule is the increase in the number of electrons from two to four (we have kept the orbital exponents equal to 1 and the internuclear distance equal to 4 a.u. unchanged). This change results in a qualitative difference in the Pauli deformation. Two large molecules For two helium atoms, the Pauli deformation means decreasing the electron den- sity in the region between the nuclei and a corresponding increase in the density on the nuclei. This looks dangerous! What if, instead of two helium atoms, we have two closed-shell long molecules A and B that touch each other with their termi- nal parts? Would the Pauli deformation be local, or would it extend over the whole system? Maybe the distant parts of the molecules would deform as much as the contact regions? Y. PAULI DEFORMATION 1057 The answer may be deduced from eq. (Y.4). The formula suggests that the elec- tronic density change pertains to the whole system. When the formula was de- rived, we concentrated on two helium atoms. However, nothing would change in the derivation if we had in mind a doubly occupied molecular orbital a that ex- tends over the whole polymer A and a similar orbital b that extends over B. In such a case the formula (Y.4) would be identical. The formula says: the three de- formation contributions cancel if we integrate them over the total space. 8 The first deformation means a density deficiency (minus sign), the other two mean density gains (plus sign). The first of these contributions is certainly located close to the con- tact region of A and B. The two others (of the same magnitude) have a spatial form such that a 2 and b 2 (i.e. extend over the whole polymer chains A and B), but are scaled by the factor 2S 2 /(1 −S 2 ). Since the contributions cancel in space (when integrated), this means that the density gain extends over the polymeric molecules and, therefore, locally is very small; the larger the interacting molecules the smaller the local change. The situation is therefore similar to an inflatable balloon pressed with your finger. We have a large deformation at the contact region , what corresponds to − 4S 1−S 2 ab, but in fact the whole balloon deforms. Because this deformation has to extend over the whole balloon, the local deformation on the other side of the toy is extremely small. Therefore, common sense has arrived at a quantum mechanical explanation. 9 This means that the Pauli deformation has a local character:ittakesplace almost exclusively in the region of contact between both molecules. Two final remarks • The Pauli deformation, treated as a spatial charge density distribution has a re- gion with positive charge (some electron density flowed from there) and negative charge (where the electron density has increased). The Pauli charge distribution participates in the Coulombic interactions within the system. If such an interac- tion is represented by a multipole–multipole interaction, the Pauli deformation has no monopole, or charge. In general, the other multipole moments of the Pauli deformation are non-zero. In particular, the Pauli deformation multipoles resulting from the exchange interaction of molecules A and B may interact with the electric multipoles of molecule C, thus contributing to the three-body effect. • If the two systems A and B approach each other in such a way that S = 0, the Pauli deformation is zero. S =0 might occur, e.g., if the two molecules approach along the nodal surfaces of the frontier molecular orbitals. 8 But of course at a given point they do not cancel in general. 9 Good for both of them. Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION In Chapter 14 the Slater determinants were constructed in three different ways using: • molecular orbitals (MO picture), • acceptor and donor orbitals (AD picture), • atomic orbitals (VB picture). Then, the problem appeared of how to express one picture by another, in partic- ular this was of importance for expressing the MO picture as an AD. More specifi- cally, we are interested in calculating the contribution of an acceptor–donor struc- ture 1 in the Slater determinant written in the MO formalism, where the molecular orbitals are expressed by the donor (n) and acceptor (χ and χ ∗ )orbitalsinthe following way ϕ 1 = a 1 n +b 1 χ −c 1 χ ∗ ϕ 2 = a 2 n −b 2 χ −c 2 χ ∗ (Z.1) ϕ 3 =−a 3 n +b 3 χ −c 3 χ ∗ We assume that {ϕ i } form an orthonormal set. For simplicity, it is also assumed that in the first approximation the orbitals {n χ χ ∗ } are also orthonormal. Then we may write that a Slater determinant in the MO picture (denoted by X i ) repre- sents a linear combination of the Slater determinants (Y j ) containing exclusively donor and acceptor orbitals: X i = j c i (Y j )Y j where the coefficient c i (Y k ) =Y k |X i at the Slater determinant Y k is the contri- bution of the acceptor–donor structure Y k in X i . In Chapter 14 three particular cases are highlighted, and they will be derived below. We will use the antisymmetrizer ˆ A = 1 N! P (−1) p ˆ P introduced in Chapter 10 ( ˆ P is the permutation operator, and p is its parity). 1 That is, of a Slater determinant built of acceptor and donor orbitals. 1058 Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION 1059 Case c 0 (DA) The c 0 (DA) coefficient means the contribution of the structure n 2 χ 2 ,i.e. (DA) =(4!) − 1 2 det[n ¯ nχ ¯χ]=(4!) 1 2 ˆ A[n ¯ nχ ¯χ] in the ground-state Slater determinant 0 =(4!) − 1 2 det[ϕ 1 ¯ϕ 1 ϕ 2 ¯ϕ 2 ]=(4!) 1 2 ˆ A[ϕ 1 ¯ϕ 1 ϕ 2 ¯ϕ 2 ] We have to calculate c 0 (DA) =Y k |X i = (DA) 0 = 4! ˆ A[n ¯ nχ ¯χ] ˆ A[ϕ 1 ¯ϕ 1 ϕ 2 ¯ϕ 2 ] = 4! [n ¯ nχ ¯χ] ˆ A 2 [ϕ 1 ¯ϕ 1 ϕ 2 ¯ϕ 2 ] = 4! [n ¯ nχ ¯χ] ˆ A[ϕ 1 ¯ϕ 1 ϕ 2 ¯ϕ 2 ] = 4! n(1) ¯ n(2)χ(3) ¯χ(4) ˆ A ϕ 1 (1) ¯ϕ 1 (2)ϕ 2 (3) ¯ϕ 2 (4) wherewehaveused ˆ A as Hermitian and idempotent. Next, we have to write all the 24 permutations [ϕ 1 (1) ¯ϕ 1 (2)ϕ 2 (3) ¯ϕ 2 (4)](taking into account their parity) and then perform integration over the coordinates of all the four electrons (together with summation over the spin variables): c 0 (DA) = dτ 1 dτ 2 dτ 3 dτ 4 n(1) ¯ n(2)χ(3) ¯χ(4) ∗ × P (−1) p P ϕ 1 (1) ¯ϕ 1 (2)ϕ 2 (3) ¯ϕ 2 (4) The integral to survive has to have perfect matching of the spin functions be- tween [ n(1) ¯ n(2)χ(3) ¯χ(4) ] and ˆ P[ϕ 1 (1) ¯ϕ 1 (2)ϕ 2 (3) ¯ϕ 2 (4)]. This makes 20 of these permutations vanish. Only four integrals will survive: c 0 (DA) = dτ 1 dτ 2 dτ 3 dτ 4 n(1) ¯ n(2)χ(3) ¯χ(4) ∗ ϕ 1 (1) ¯ϕ 1 (2)ϕ 2 (3) ¯ϕ 2 (4) − dτ 1 dτ 2 dτ 3 dτ 4 n(1) ¯ n(2)χ(3) ¯χ(4) ∗ ϕ 1 (1) ¯ϕ 1 (4)ϕ 2 (3) ¯ϕ 2 (2) − dτ 1 dτ 2 dτ 3 dτ 4 n(1) ¯ n(2)χ(3) ¯χ(4) ∗ ϕ 1 (3) ¯ϕ 1 (2)ϕ 2 (1) ¯ϕ 2 (4) + dτ 1 dτ 2 dτ 3 dτ 4 n(1) ¯ n(2)χ(3) ¯χ(4) ∗ ϕ 1 (3) ¯ϕ 1 (4)ϕ 2 (1) ¯ϕ 2 (2) 1060 Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION = dτ 1 n(1) ∗ ϕ 1 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 1 (2) dτ 3 χ(3) ∗ ϕ 2 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 2 (4) − dτ 1 n(1) ∗ ϕ 1 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 2 (2) dτ 3 χ(3) ∗ ϕ 2 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 1 (4) − dτ 1 n(1) ∗ ϕ 2 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 1 (2) dτ 3 χ(3) ∗ ϕ 1 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 2 (4) + dτ 1 n(1) ∗ ϕ 2 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 2 (2) dτ 3 χ(3) ∗ ϕ 1 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 1 (4) =(a 1 ) 2 (−b 2 ) 2 −a 1 a 2 (−b 2 )b 1 −a 2 a 1 b 1 (−b 2 ) +(a 2 ) 2 (b 1 ) 2 =(a 1 ) 2 (b 2 ) 2 +a 1 a 2 b 2 b 1 +a 2 a 1 b 1 b 2 +(a 2 ) 2 (b 1 ) 2 =a 1 b 2 (a 1 b 2 +a 2 b 1 ) +a 2 b 1 (a 1 b 2 +a 2 b 1 ) =(a 1 b 2 +a 2 b 1 ) 2 = a 1 a 2 b 1 −b 2 2 Hence, c 0 (DA) = a 1 a 2 b 1 −b 2 2 which agrees with the formula on p. 805. Case c 2 (DA) The c 2 (DA) represents the contribution of the structure (DA) =(4!) 1 2 ˆ A[n ¯ nχ ¯χ] in the Slater determinant corresponding to the double excitation 2d = (4!) 1 2 ˆ A[ϕ 1 ¯ϕ 1 ϕ 3 ¯ϕ 3 ]. We are interested in the integral c 2 (DA) = (DA) 2d = 4! n(1) ¯ n(2)χ(3) ¯χ(4) ˆ A ϕ 1 (1) ¯ϕ 1 (2)ϕ 3 (3) ¯ϕ 3 (4) This case is very similar to the previous one, the only difference is the substitu- tion ϕ 2 →ϕ 3 . Therefore, everything goes the same way as before, but this time we obtain: c 2 (DA) = dτ 1 n(1) ∗ ϕ 1 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 1 (2) dτ 3 χ(3) ∗ ϕ 3 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 3 (4) − dτ 1 n(1) ∗ ϕ 1 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 3 (2) dτ 3 χ(3) ∗ ϕ 3 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 1 (4) − dτ 1 n(1) ∗ ϕ 3 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 1 (2) dτ 3 χ(3) ∗ ϕ 1 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 3 (4) Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION 1061 + dτ 1 n(1) ∗ ϕ 3 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 3 (2) dτ 3 χ(3) ∗ ϕ 1 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 1 (4) or c 2 (DA) = (a 1 ) 2 (b 3 ) 2 −a 1 (−a 3 )b 3 b 1 −(−a 3 )a 1 b 1 b 3 +(−a 3 ) 2 (b 1 ) 2 = (a 1 ) 2 (b 3 ) 2 +a 1 a 3 b 3 b 1 +a 3 a 1 b 1 b 3 +(a 3 ) 2 (b 1 ) 2 =(a 1 b 3 +a 3 b 1 ) 2 = a 1 b 1 −a 3 b 3 2 We have c 2 (DA) = a 1 b 1 −a 3 b 3 2 which also agrees with the result used on p. 806. Case c 3 (DA) This time we have to calculate the contribution of (DA) =(4!) 1 2 ˆ A[n ¯ nχ ¯χ] in the Slater determinant 3d =(4!) 1 2 ˆ A[ϕ 2 ¯ϕ 2 ϕ 3 ¯ϕ 3 ], therefore c 2 (DA) = (DA) 3d = 4! n(1) ¯ n(2)χ(3) ¯χ(4) ˆ A ϕ 2 (1) ¯ϕ 2 (2)ϕ 3 (3) ¯ϕ 3 (4) This is a similar case to the previous one, but we have to exchange ϕ 1 →ϕ 2 .We obtain: c 3 (DA) = dτ 1 n(1) ∗ ϕ 2 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 2 (2) dτ 3 χ(3) ∗ ϕ 3 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 3 (4) − dτ 1 n(1) ∗ ϕ 2 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 3 (2) dτ 3 χ(3) ∗ ϕ 3 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 2 (4) − dτ 1 n(1) ∗ ϕ 3 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 2 (2) dτ 3 χ(3) ∗ ϕ 2 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 3 (4) + dτ 1 n(1) ∗ ϕ 3 (1) dτ 2 ¯ n(2) ∗ ¯ϕ 3 (2) dτ 3 χ(3) ∗ ϕ 2 (3) dτ 4 ¯χ(4) ∗ ¯ϕ 2 (4) or c 3 (DA) = (a 2 ) 2 (b 3 ) 2 −a 2 (−a 3 )b 3 (−b 2 ) −(−a 3 )a 2 (−b 2 )b 3 +(−a 3 ) 2 (−b 2 ) 2 = (a 2 ) 2 (b 3 ) 2 −a 2 a 3 b 3 b 2 −a 3 a 2 b 2 b 3 +(a 3 ) 2 (b 2 ) 2 = a 2 b 3 [a 2 b 3 −a 3 b 2 ]−a 3 b 2 [a 2 b 3 −a 3 b 2 ] = (a 2 b 3 −a 3 b 2 ) 2 = a 2 −b 2 −a 3 b 3 2 1062 Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION Finally, c 3 (DA) = a 2 −b 2 −a 3 b 3 2 and again agreement with the formula on p. 806 is obtained. Table Z.1. Units of physical quantities Quantity Unit Symbol Value light velocity c 299792458 km s Planck constant h 66260755 ·10 −34 J ·s mass electron rest mass m 0 91093897 ·10 −31 kg charge element. charge = a.u. of charge e 160217733 ·10 −19 C action h 2π ¯ h 105457266 ·10 −34 J ·s length bohr = a.u. of length a 0 529177249 ·10 −11 m energy hartree = a.u. of energy E h 43597482 ·10 −18 J time a.u. of time ¯ h E h 2418884 ·10 −17 s velocity a.u. of velocity a 0 E h ¯ h 2187691 ·10 6 m s momentum a.u. of momentum ¯ h a 0 1992853 ·10 −24 kgm s electr. dipole moment a.u. of electr. dipole ea 0 8478358 ·10 −30 C ·m (2.5415 D) magn. dipole Bohr magneton e ¯ h 2m 0 c 092731 ·10 −20 erg gauss polarizability e 2 a 2 0 E h 1648778 ·10 −41 C 2 m 2 J electric field E h ea 0 5142208 ·10 11 V m Boltzm. constant k B 1380658 ·10 −23 J K Avogadro constant N A 60221367 ·10 23 mol −1 Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION 1063 Table Z.2. Conversion coefficients a.u. erg eV kcal mole 1cm −1 1Hz 1K 1a.u. 1 435916 ·10 −11 272097 627709 2194746 ·10 5 6579695 ·10 15 315780 ·10 5 1erg 229402 ·10 10 1624197 ·10 11 143998 ·10 13 503480 ·10 15 150940 ·10 26 72441 ·10 15 1eV 367516 ·10 −2 160206 ·10 −12 1230693 806604 ·10 3 241814 ·10 14 116054 ·10 4 1 kcal mol 159310 ·10 −3 69446 ·10 −14 433477 ·10 −2 1349644 ·10 2 1048209 ·10 13 50307 ·10 2 1cm −1 4556336 ·10 −6 198618 ·10 −16 123977 ·10 −4 286005 ·10 −3 12997930 ·10 10 143880 1Hz 1519827 ·10 −16 662517 ·10 −27 413541 ·10 −15 954009 ·10 −14 3335635 ·10 −11 147993 ·10 −11 1K 316676 ·10 −6 138044 ·10 −16 86167 ·10 −5 198780 ·10 −3 069502 208363 ·10 10 1 This page intentionally left blank N AME I NDEX Abragam Anatole 668 Abramovitz Milton 481 Adamowicz Ludwik 513, 573 Adams John E. 765, 784, 790 Adleman Leonard M. 851, 878, 879, 880 Aharonov Yakir 968 Ahlrichs Reinhart 356, 532 Albrecht Andreas A. 395 Alder Berni Julian 278, 825 Alderton Mark 702, 1019 Alexander Steven 269 Alijah Alexander 222, 272, 273 Allen Michael P. 322 Allinger Norman L. 286, 291 Amos A. Terry 716 Anderson Carl David 14, 113, 268 André Jean-Marie 90, 140, 374, 431, 465, 487, 496, 642, 644 André Marie-Claude 90, 140 Andzelm Jan 602, 612 Anfinsen Christian 294 Aquilanti Vincenzo 742 Arndt Markus 43 Arrhenius Svante August 832 Aspect Alain 3, 14, 46, 53, 54 Atkins Peter William 70, 381, 920 Auger Pierre Victor 270 van der Avoird Ad 284, 717 Axilrod Benjamin M. 565, 741, 758, 761 Ayers Paul W. 395 Babloyantz Agnes 885 Bader Richard F.W. 87, 418, 509, 569, 571, 572, 575, 576–578, 608, 612, 708, 767 Baerends Evert Jan 612 Bak Keld L. 678 Baker Jon 602 Bała Piotr 309 Balakrishnan Ashok 510 Balmer Johann Jacob 6 Banach Stefan 219, 311 Barbara Paul F. 830 Barlow William 920 Barnwell John D. 765 Bartlett Rodney J. 547, 564 Barysz Maria 118, 138 Bautista Debra L. 749 Bayly Christopher I. 288, 745 Becke Axel D. 591, 603, 612, 613 Bell John 3, 14, 43, 45, 46, 53, 54 Belousov Boris Pavlovich 850 Bennett Charles H. 3, 47, 53 Bernardi Fernando 262, 273, 690 Bernstein Richard B. 766, 845 Berry Michael V. 264, 265, 764, 780, 781 Berthier Gaston 397, 400, 536 Bethe Hans Albrecht 131, 140 Beutler Hans 510 Bielejewska Anna 856 Bie ´ nko Dariusz C. 559 Binkley J. Stephen 547 Bird R. Byron 760 Bishop David M. 510 Bixon Mordechai 277, 289 Bjorken James D. 111 Bloch Claude 714 Bloch Felix 328, 331, 360, 361, 431, 434, 435, 439, 447–452, 462, 493, 494, 496, 663 Blume Doerte 775 Boese Roland 388 Bogolyubov Nicolai Nicolaevich 492, 493 1065 . ·10 −18 J time a.u. of time ¯ h E h 2418884 ·10 −17 s velocity a.u. of velocity a 0 E h ¯ h 2187691 ·10 6 m s momentum a.u. of momentum ¯ h a 0 1992853 ·10 −24 kgm s electr. dipole moment a.u. of electr of the Pauli deformation are non-zero. In particular, the Pauli deformation multipoles resulting from the exchange interaction of molecules A and B may interact with the electric multipoles of. molecules approach along the nodal surfaces of the frontier molecular orbitals. 8 But of course at a given point they do not cancel in general. 9 Good for both of them. Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS