Ideas of Quantum Chemistry P110

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 −S2)ab in the contact region, 2S2/(1 − S2)a2 on molecule A and 2S2/(1 − S2)b2 on polymer B – this is why 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 2S2/(1−S2)a2 for molecule A has been added to the initial density distribution, and similarly for molecule B (the rectangles became larger, but locally the corresponding increaseissmall). (d) In order to represent the total Pauli deformation from the result obtained at point c we subtracted the density distribution 4S/(1−S2)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 certainlylocatedclosetothecon-tact region of A and B. The two others (of the same magnitude) have a spatial form such that a2 and b2 (i.e. extend over the whole polymer chains A and B), but are scaled by the factor 2S2/(1 − S2). 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 localchange. 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 − 4S2 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: it takes place 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-gionwithpositivecharge(someelectrondensityflowedfromthere)andnegative 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. 8But of course at a given point they do not cancel in general. 9Good 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-ture1 in the Slater determinant written in the MO formalism, where the molecular orbitals are expressed by the donor (n) and acceptor (χ and χ∗) orbitals in the following way ϕ1 = a1n+b1χ−c1χ∗; ϕ2 = a2n−b2χ−c2χ∗; (Z.1) ϕ3 = −a3n+b3χ−c3χ∗: We assume that {ϕi} 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 Xi) repre-sents a linear combination of the Slater determinants (Yj) containing exclusively donor and acceptor orbitals: X Xi = ci(Yj)Yj; j where the coefficient ci(Yk)=hYk|Xii at the Slater determinant Yk is the contri-bution of the acceptor–donor structure Yk in Xi. In Chapter 14 three particular cases are highlighted, and they will be derived below. We will use the antisymmetrizer A= N! X(−1)pP introduced in Chapter 10 (P is the permutation operator, and p is its parity). 1That is, of a Slater determinant built of acceptor and donor orbitals. 1058 Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION 1059 Case c0(DA) The c0(DA) coefficient means the contribution of the structure n2χ2, i.e. 9(DA)=(4!)−2 det[nnχχ¯]=(4!)2 A[nnχχ¯] in the ground-state Slater determinant 90 =(4!)−2 det[ϕ1ϕ¯1ϕ2ϕ¯2]=(4!)2 A[ϕ1ϕ¯1ϕ2ϕ¯2]: We have to calculate c0(DA) = hYk|Xii=­9(DA)¯90® = 4! A[nnχχ¯]¯A[ϕ1ϕ¯1ϕ2ϕ¯2] = 4!­[nnχχ¯]¯ ˆ2[ϕ1ϕ¯1ϕ2ϕ¯2]® = 4!­[nnχχ¯]¯A[ϕ1ϕ¯1ϕ2ϕ¯2]® = 4!­£n(1)n(2)χ(3)χ¯(4)¤¯A£ϕ1(1)ϕ¯1(2)ϕ2(3)ϕ¯2(4)¤®; where we have used A as Hermitian and idempotent. Next, we have to write all the 24 permutations [ϕ1(1)ϕ¯1(2)ϕ2(3)ϕ¯2(4)](takinginto accounttheir parity)and then perform integration over the coordinates of all the four electrons (together with summation over the spin variables): c0(DA) = Z dτ1 dτ2 dτ3 dτ4 £n(1)n(2)χ(3)χ¯(4)¤∗ ×X(−1)pP£ϕ1(1)ϕ¯1(2)ϕ2(3)ϕ¯2(4)¤: P The integral to survive has to have perfect matching of the spin functions be- tween [n(1)n(2)χ(3)χ¯(4)] and P[ϕ1(1)ϕ¯1(2)ϕ2(3)ϕ¯2(4)]. This makes 20 of these permutations vanish. Only four integrals will survive: c0(DA) = dτ1 dτ2 dτ3 dτ4 £n(1)n(2)χ(3)χ¯(4)¤∗£ϕ1(1)ϕ¯1(2)ϕ2(3)ϕ¯2(4)¤ −Z dτ1 dτ2 dτ3 dτ4 £n(1)n(2)χ(3)χ¯(4)¤∗£ϕ1(1)ϕ¯1(4)ϕ2(3)ϕ¯2(2)¤ −Z dτ1 dτ2 dτ3 dτ4 £n(1)n(2)χ(3)χ¯(4)¤∗£ϕ1(3)ϕ¯1(2)ϕ2(1)ϕ¯2(4)¤ +Z dτ1 dτ2 dτ3 dτ4 £n(1)n(2)χ(3)χ¯(4)¤∗£ϕ1(3)ϕ¯1(4)ϕ2(1)ϕ¯2(2)¤ 1060 Z. ACCEPTOR–DONOR STRUCTURE CONTRIBUTIONS IN THE MO CONFIGURATION Z Z Z Z = dτ1 n(1)∗ϕ1(1) dτ2 n(2)∗ϕ¯1(2) dτ3 χ(3)∗ϕ2(3) dτ4 χ¯(4)∗ϕ¯2(4) Z Z Z Z − dτ1 n(1)∗ϕ1(1) dτ2 n(2)∗ϕ¯2(2) dτ3 χ(3)∗ϕ2(3) dτ4 χ¯(4)∗ϕ¯1(4) Z Z Z Z − dτ1 n(1)∗ϕ2(1) dτ2 n(2)∗ϕ¯1(2) dτ3 χ(3)∗ϕ1(3) dτ4 χ¯(4)∗ϕ¯2(4) Z Z Z Z + dτ1 n(1)∗ϕ2(1) dτ2 n(2)∗ϕ¯2(2) dτ3 χ(3)∗ϕ1(3) dτ4 χ¯(4)∗ϕ¯1(4) =(a1)2(−b2)2 −a1a2(−b2)b1 −a2a1b1(−b2)+(a2)2(b1)2 =(a1)2(b2)2 +a1a2b2b1 +a2a1b1b2 +(a2)2(b1)2 =a1b2(a1b2 +a2b1)+a2b1(a1b2 +a2b1) ¯ ¯2 =(a1b2 +a2b1)2 =¯b1 −b2 ¯ : Hence, ¯ c0(DA)=¯b1 a ¯2 −b2 ¯ which agrees with the formula on p. 805. Case c2(DA) The c2(DA) represents the contribution of the structure 9(DA)=(4!)1 A[nnχχ¯] in the Slater determinant corresponding to the double excitation 92d = (4!)1 A[ϕ1ϕ¯1ϕ3ϕ¯3]. We are interested in the integral c2(DA) = ­9(DA)¯92d® = 4!­£n(1)n(2)χ(3)χ¯(4)¤¯A£ϕ1(1)ϕ¯1(2)ϕ3(3)ϕ¯3(4)¤®: This case is very similar to the previous one, the only difference is the substitu-tion ϕ2 →ϕ3. Therefore, everything goes the same way as before, but this time we obtain: c2(DA) Z Z Z = dτ1 n(1)∗ϕ1(1) dτ2 n(2)∗ϕ¯1(2) dτ3 χ(3)∗ϕ3(3) dτ4 χ¯(4)∗ϕ¯3(4) Z Z Z Z − dτ1 n(1)∗ϕ1(1) dτ2 n(2)∗ϕ¯3(2) dτ3 χ(3)∗ϕ3(3) dτ4 χ¯(4)∗ϕ¯1(4) Z Z Z Z − dτ1 n(1)∗ϕ3(1) dτ2 n(2)∗ϕ¯1(2) dτ3 χ(3)∗ϕ1(3) dτ4 χ¯(4)∗ϕ¯3(4) ... - tailieumienphi.vn