1026 U. SECOND QUANTIZATION One-electron operators The operator ˆ F = i ˆ h(i) is the sum of the one-electron operators 7 ˆ h(i) acting on functions of the coordinates of electron i. The I Slater–Condon rule says (see Appendix M), that for the Slater determi- nant ψ built of the spinorbitals φ i , the matrix element ψ| ˆ Fψ= i h ii ,where h ij =φ i | ˆ hφ j . In the second quantization ˆ F = ∞ ij h ij ˆ i † ˆ j Interestingly, the summation extends to infinity, and therefore the operator is independent of the number of electrons in the system. Let us check whether the formula is correct. Let us insert ˆ F = ij h ij ˆ ı † ˆ into ψ| ˆ Fψ.Wehave ψ ˆ Fψ = ψ ij h ij ˆ ı † ˆ ψ = ij h ij ψ ˆ ı † ˆ ψ = ij h ij δ ij = i h ii This is correct. What about the II Slater–Condon rule (the Slater determinants ψ 1 and ψ 2 differ by a single spinorbital: the spinorbital i in ψ 1 is replaced by the spinorbital i in ψ 2 )? We have ψ 1 ˆ Fψ 2 = ij h ij ψ 1 ˆ ı † ˆ ψ 2 The Slater determinants that differ by one spinorbital produce an overlap inte- gral equal to zero, 8 therefore ψ 1 | ˆ Fψ 2 =h ii . Thus, the operator in the form ˆ F = ij h ij ˆ ı † ˆ ensures equivalence with all the Slater–Condon rules. Two-electron operators Similarly, we may use the creation and annihilation operators to represent the two- electron operators ˆ G = 1 2 ij ˆ g(ij). In most cases ˆ g(ij) = 1 r ij and ˆ G has the form: ˆ G = 1 2 ij 1 r ij = 1 2 ∞ ij kl ij |kl ˆ j † ˆ i † ˆ k ˆ l 7 Most often this will be the kinetic energy operator, the nuclear attraction operator, the interaction with the external field or the multipole moment. 8 It is evident, that if in this situation the Slater determinants ψ 1 and ψ 2 differed by more than a single spinorbital, we would get zero (III and IV Slater–Condon rule). U. SECOND QUANTIZATION 1027 Here also the summation extends to infinity and the operator is independent of the number of electrons in the system. The proof of the I Slater–Condon rule relies on the following chain of equalities ψ ˆ Gψ = 1 2 ij kl ij |kl ψ ˆ † ˆ ı † ˆ k ˆ lψ = 1 2 ij kl ij |kl ˆ ı ˆ ψ ˆ k ˆ lψ = 1 2 ij kl ij |kl(δ ik δ jl −δ il δ jk ) = 1 2 ij ij |ij −ij |ji because the overlap integral ˆ ı ˆ ψ| ˆ k ˆ lψof the two Slater determinants ˆ ı ˆ ψ and ˆ k ˆ lψ is non-zero in the two cases only: either if i = k, j = l or if i = l, j = k (then the sign has to change). This is what we get from the Slater–Condon rules. For the II Slater–Condon rule we have (instead of the spinorbital i in ψ 1 we have the spinorbital i in ψ 2 ): ψ 1 ˆ Gψ 2 = 1 2 Ijkl Ij|kl ψ 1 ˆ † ˆ I † ˆ k ˆ lψ 2 = 1 2 Ijkl Ij|kl ˆ I ˆ ψ 1 ˆ k ˆ lψ 2 (U.1) where the summation index I has been introduced in order not to mix with spinor- bital i. In the overlap integral ˆ I ˆ ψ 1 | ˆ k ˆ lψ 2 the sets of the spinorbitals in the Slater determinant ˆ I ˆ ψ 1 and in the Slater determinant ˆ k ˆ lψ 2 have to be identical,other- wise the integral will equal zero. However, in ψ 1 and ψ 2 we already have a differ- ence of one spinorbital. Thus, first we have to get rid of these spinorbitals (i and i ). For the integral to survive 9 we have to have at least one of the following conditions satisfied: • I =i and k =i (and then j =l), • j =i and k =i (and then I =l), • I =i and l =i (and then j =k), • j =i and l =i (and then I =k). This means that, taking into account the above cases in eq. (U.1), we obtain ψ 1 ˆ Gψ 2 = 1 2 j ij i j ˆ ı ˆ ψ 1 ˆ ı ˆ ψ 2 + 1 2 l li i l ˆ l ˆ ıψ 1 ˆ ı ˆ lψ 2 + 1 2 j ij ji ˆ ı ˆ ψ 1 ˆ ˆ ı ψ 2 + 1 2 k ki ki ˆ k ˆ ıψ 1 ˆ k ˆ ı ψ 2 = 1 2 j ij i j − 1 2 l li i l − 1 2 j ij ji + 1 2 k ki ki = 1 2 j ij i j − 1 2 j ji i j − 1 2 j ij ji + 1 2 j ji ji 9 This is a necessary, but not a sufficient condition. 1028 U. SECOND QUANTIZATION = 1 2 j ij i j − 1 2 j ij ji − 1 2 j ij ji + 1 2 j ij i j = j ij i j − j ij ji where the coordinates of electrons 1 and 2 have been exchanged in the two sums. Notice that the overlap integrals ˆ ı ˆ ψ 1 ˆ ı ˆ ψ 2 = ˆ k ˆ ıψ 1 ˆ k ˆ ı ψ 2 =1 because the Slater determinants ˆ iψ 1 and ˆ i ψ 2 are identical. Also, from the anti- commutation rules ˆ l ˆ ıψ 1 ˆ ı ˆ lψ 2 = ˆ ı ˆ ψ 1 ˆ ˆ ı ψ 2 =−1 Thus the II Slater–Condon rule has been correctly reproduced: ψ 1 ˆ Gψ 2 = j ij i j − ij ji We may conclude that the definition of the creation and annihilation operators and the simple anticommutation relations are equivalent to the Slater–Condon rules. This opens up the space spanned by the Slater determinants for us, i.e. all the integrals involving Slater determinants can be easily transformed into one- and two-electron integrals involving spinorbitals. V. THE HYDROGEN ATOM IN THE ELECTRIC FIELD – VARIATIONAL APPROACH Polarization of an atom or molecule can be calculated by using the finite field method described on p. 639. Let us apply this method to the hydrogen atom. Its polarizability was already calculated using a simple version of perturbation theory (p. 636). This time we will use the variational method. The Hamiltonian for the isolated hydrogen atom (within the Born–Oppenhei- mer approximation) reads as ˆ H (0) =− 1 2 e − 1 r where the first term is the electronic kinetic energy operator, and the second its Coulomb interaction energy with the nucleus (proton–electron distance is denoted by r). The atom is in a homogeneous electric field E =(0 0 E) with E > 0andas in perturbation theory (p. 636), the total Hamiltonian has the form ˆ H = ˆ H (0) +V with V =zE,wherez denotes the coordinate of the electron and the proton is at the origin (the derivation of the formula is given on p. 636, the exchange of z to x does not matter). The variational wave function ψ is proposed in the form ψ =χ 1 +cχ 2 (V.1) where χ 1 = 1 √ π exp(−r) is the 1s orbital of the hydrogen atom (ground state) and χ 2 is the normalized 1 p-type orbital χ 2 =Nzexp(−ζr) 1 N can be easily calculated from the normalization condition 1 = N 2 z exp(−ζr) 2 dV =N 2 ∞ 0 drr 4 exp(−2ζr) π 0 dθ sinθ cos 2 θ 2π 0 dφ = N 2 4!(2ζ) −5 2 3 2π = N 2 π ζ 5 This gives N = ζ 5 π . 1029 1030 V. THE HYDROGEN ATOM IN THE ELECTRIC FIELD – VARIATIONAL APPROACH There are two variational parameters c and ζ. Let us assume for a while that we have fixed the value of ζ, so the only variational parameter is c.Thewave function ψ is a linear combination of two expansion functions (“two-state model”): χ 1 and χ 2 . Therefore, optimal energy follows from the Ritz method, according to case III of Appendix D on p. 948: E =E ar ± 2 +h 2 (V.2) where arithmetic mean energy E ar ≡ H 11 +H 22 2 ,while ≡ H 11 −H 22 2 and h ≡ H 12 = H 21 with H ij ≡ χ i ˆ Hχ j = χ i ˆ H (0) χ j +χ i |Vχ j Let us calculate all the ingredients of the energy given by (V.2). First, let us note that H 11 ≡χ 1 | ˆ H (0) χ 1 =− 1 2 a.u., since χ 1 is the ground state of the isolated hydrogen atom (p. 178), and V 11 =χ 1 |Vχ 1 =0, because the inte- grand is antisymmetric with respect to z →−z. Now let us calculate H 22 = H (0) 22 +V 22 .NotethatV 22 = 0, for the same reason as V 11 .Wehave H (0) 22 =− 1 2 χ 2 | e χ 2 − χ 2 1 r χ 2 The second integral is χ 2 1 r χ 2 = N 2 ∞ 0 drr 3 exp(−2ζr) π 0 dθ sinθ cos 2 θ 2π 0 dφ = ζ 5 π ·3!(2ζ) −4 · 2 3 ·2π = 1 2 ζ where the dots separate the values of the corresponding integrals. 2 In Appendix R, the reader will find the main ingredients needed to calculate the first integral of H (0) 22 : χ 2 | e χ 2 =N 2 r cosθ exp(−ζr) 1 r 2 ∂ ∂r r 2 ∂ ∂r + 1 r 2 sinθ ∂ ∂θ sinθ ∂ ∂θ + 1 r 2 sin 2 θ ∂ 2 ∂φ 2 r cosθ exp(−ζr) = N 2 r cosθ exp(−ζr) cosθ 1 r 2 ∂ ∂r r 2 exp(−ζr) −ζr 3 exp(−ζr) + r cosθ exp(−ζr) (−2cosθ) r 2 r exp(−ζr) +0 = N 2 r cosθ exp(−ζr) cosθ 2 r −ζ −3ζ +ζ 2 r exp(−ζr) 2 Note that, in spherical coordinates, the volume element dV = r 2 sinθ dr dθ dφ. In derivations of this Appendix (and not only) we often use the equality ∞ 0 dxx n exp(−αr) =n!α −(n+1) . V. THE HYDROGEN ATOM IN THE ELECTRIC FIELD – VARIATIONAL APPROACH 1031 + r cosθ exp(−ζr) (−2cosθ) r exp(−ζr) = ζ 5 π 2 3 ·2π 2 ·2 (2ζ) 3 − 4ζ ·3! (2ζ) 4 + ζ 2 ·4! (2ζ) 5 − 2 ·2! (2ζ) 3 =−ζ 2 Thus, we obtain H 22 = 1 2 ζ 2 − 1 2 ζ. This formula looks good, since for χ 2 =2p z , i.e. for ζ = 1 2 we get correctly (see p. 178) H 22 = E 2p =− 1 8 a.u., the energy of orbital 2p. Let us turn to the non-diagonal matrix element of the Hamiltonian: H 12 = H (0) 12 + V 12 . Note, that H (0) 12 = 0, because χ 1 is an eigenfunction of ˆ H (0) and χ 1 |χ 2 =0. Thus, h =NE r cosθ exp(−ζ) r cosθ 1 √ π exp(−r) = NE 1 √ π ∞ 0 drr 4 exp −(ζ +1)r π 0 dθ sinθ cos 2 θ 2π 0 dφ = E ζ 5 π ·4!(ζ +1) −5 · 2 3 ·2π =32 ζ 5 (ζ +1) 5 E Now we can write eq. (V.2) as a function of ζ: E = 1 4 ζ 2 −ζ −1 − 1 16 ζ 2 −ζ +1 2 +ζ 5 2 ζ +1 10 E 2 (V.3) We would like to expand this expression in a power series of E to highlight the coefficient at E 2 , because this coefficient is related to the polarizability. The expansion gives (in a.u.) E ≈ 1 4 ζ 2 −ζ −1 − 1 4 ζ 2 −ζ +1 − 1 2 α zz E 2 +···=− 1 2 − 1 2 α zz E 2 +··· where, according to eq. (12.24), the polarizability (in a.u.) reads as α zz =4 · ζ 5 |ζ 2 −ζ +1| 2 ζ +1 10 (V.4) Several numerical values of α zz calculated using (V.3) and (V.4), are given on p. 639. They may be compared with the exact result α zz =45a.u. W. NMR SHIELDING AND COUPLING CONSTANTS – DERIVATION This section is for those who do not fully believe the author, and want to check whether the final formulae for the shielding and coupling constants in nuclear mag- netic resonance are indeed valid (Chapter 12). 1 SHIELDING CONSTANTS Let us begin with eq. (12.87). Applying vector identities We are going to apply some vector identities 1 in the operators ˆ B 3 ˆ B 4 ˆ B 5 .The first identity is u ·(v ×w) =v · (w ×u) =w · (u × v), which simply means three equivalent ways of calculating the volume of a parallelepiped (cf. p. 437). This identity applied to ˆ B 3 and ˆ B 4 gives ˆ B 3 = e mc A j γ A I A · ˆ L Aj r 3 Aj (W.1) ˆ B 4 = e 2mc j H · ˆ L 0j (W.2) Let us transform the term ˆ B 5 by using the following identity (u ×v) ·(w ×s) = (u ·w)(v ·s) −(v ·w)(u ·s): ˆ B 5 = e 2 2mc 2 A j γ A (H ×r 0j ) · I A ×r Aj r 3 Aj = e 2 2mc 2 A j γ A (H ·I A )(r 0j ·r Aj ) −(r 0j ·I A )(H ·r Aj ) · 1 r 3 Aj Putting things together Now we are all set to put all this baroque furniture into its place, i.e. into eq. (12.87) for E 1 The reader may easily check each of them. 1032 1 Shielding constants 1033 E = A E A (W.3) where E A stands for the contribution of nucleus A: E A =−γ A ψ (0) 0 (I A ·H)ψ (0) 0 + e 2 2mc 2 γ A ψ (0) 0 j (H ·I A )(r 0j ·r Aj ) −(r 0j ·I A )(H ·r Aj ) · 1 r 3 Aj ψ (0) 0 + e 2 2m 2 c 2 γ A ψ (0) 0 j I A · ˆ L Aj r 3 Aj ˆ R 0 j H · ˆ L 0j ψ (0) 0 + ψ (0) 0 j H · ˆ L 0j ˆ R 0 j I A · ˆ L Aj r 3 Aj ψ (0) 0 Averaging over rotations The expression for E A represents a bilinear form with respect to the components of vectors I A and H E A =I T A C A H where C A stands for a square matrix 2 of dimension 3, and I A and H are vertical three-component vectors. A contribution to the energy such as E A cannot depend on our choice of co- ordinate system axes x yz, i.e. on the components of I A and H.Wewillobtain the same energy if we rotate the axes (orthogonal transformation) in such a way as to diagonalize C A . The resulting diagonalized matrix C Adiag has three eigen- values (composing the diagonal) corresponding to the new axes x y z . The very essence of averaging is that none of these axes are to be privileged in any sense. This is achieved by constructing the averaged matrix 1 3 (C Adiag ) x x +(C Adiag ) y y +(C Adiag ) z z = ¯ C Adiag x x = ¯ C Adiag y y = ¯ C Adiag z z ≡C A where ( ¯ C Adiag ) qq = δ qq C A for q q = x y z Note that since the transforma- tion was orthogonal (i.e. the trace of the matrix is preserved), the number C A may also be obtained from the original matrix C A C A = 1 3 (C Adiag ) x x +(C Adiag ) y y +(C Adiag ) z z = 1 3 [C Axx +C Ayy +C Azz ] (W.4) 2 We could write its elements from equation for E A , but their general form will turn out to be not necessary. 1034 W. NMR SHIELDING AND COUPLING CONSTANTS – DERIVATION Then the averaged energy E becomes (note the resulting dot product) ¯ E = A I T A ¯ C Adiag H = A C A (I A ·H) Thus we obtain the sum of energy contributions over the nuclei, each contribu- tion with its own coefficient averaged over rotations 3 ¯ E =− A γ A I A ·H 1 − e 2 2mc 2 ψ (0) 0 j 2 3 (r 0j ·r Aj ) 1 r 3 Aj ψ (0) 0 − e 2 2m 2 c 2 1 3 ψ (0) 0 j ˆ L Aj r 3 Aj ˆ R 0 j ˆ L 0j + j ˆ L 0j ˆ R 0 j ˆ L Aj r 3 Aj ψ (0) 0 (W.5) with the matrix elements ˆ U kl = ψ (0) k ˆ Uψ (0) l of the corresponding operators ˆ U =( ˆ U x ˆ U y ˆ U z ). Finally, after comparing the formula with eq. (12.80), we obtain the shielding constant for nucleus A (the change of sign in the second part of the formula comes from the change in the denominator) given in eq. (12.88). 3 Indeed, making C A = 1 3 [C Axx +C Ayy +C Azz ] for the terms of eq. (W.3) we have the following contributions (term by term): •−γ A 1 3 [ 1 +1 +1 ] =−γ A ; • e 2 2mc 2 γ A 1 3 ψ (0) 0 j r 0j ·r Aj 1 r 3 Aj ψ (0) 0 + ψ (0) 0 j r 0j ·r Aj 1 r 3 Aj ψ (0) 0 + ψ (0) 0 j r 0j ·r Aj 1 r 3 Aj ψ (0) 0 = e 2 2mc 2 γ A ψ (0) 0 j r 0j ·r Aj 1 r 3 Aj ψ (0) 0 ; •− e 2 2mc 2 γ A ψ (0) 0 j 1 3 [x 0j x Aj +y 0j y Aj +z 0j z Aj ] 1 r 3 Aj ψ (0) 0 =− e 2 2mc 2 γ A ψ (0) 0 j 1 3 r 0j ·r Aj 1 r 3 Aj ψ (0) 0 + 1 3 e 2 2m 2 c 2 γ A k 1 E (0) 0 −E (0) k × ψ (0) 0 j ˆ L Ajx r 3 Aj ψ (0) k ψ (0) k j ˆ L 0jx ψ (0) 0 +similarly yz +cc = 1 3 e 2 2m 2 c 2 γ A k 1 E (0) 0 −E (0) k × 1 3 ψ (0) 0 j ˆ L Aj r 3 Aj ψ (0) k ψ (0) k j ˆ L 0j ψ (0) 0 +cc where cc means the “complex conjugate” counterpart. This reproduces eq. (W.5). 2 Coupling constants 1035 2 COUPLING CONSTANTS Averaging over rotations In each contribution on p. 670 there is a double summation over the nuclear spins, which, after averaging over rotations (as for the shielding constant) gives rise to an energy dependence of the kind A<B γ A γ B K AB ˆ I A · ˆ I B which is required in the NMR Hamiltonian. Now, let us take the terms E DSO , E PSO , E SD , E FC and average them over rotations producing ¯ E DSO , ¯ E PSO , ¯ E SD , ¯ E FC : • ¯ E DSO = e 2 2mc 2 AB j γ A γ B I A ·I B ψ (0) 0 r Aj ·r Bj r 3 Aj r 3 Bj ψ (0) 0 − e 2 2mc 2 AB j γ A γ B 1 3 I A ·I B × ψ (0) 0 x Aj x Bj r 3 Aj r 3 Bj ψ (0) 0 + ψ (0) 0 y Aj y Bj r 3 Aj r 3 Bj ψ (0) 0 + ψ (0) 0 z Aj z Bj r 3 Aj r 3 Bj ψ (0) 0 because the first part of the formula does not need any averaging (it is already in the appropriate form), the second part is averaged according to (W.4). Therefore, ¯ E DSO = e 2 3mc 2 AB j γ A γ B I A ·I B ψ (0) 0 r Aj ·r Bj r 3 Aj r 3 Bj ψ (0) 0 • ¯ E PSO = ψ (0) 0 ˆ B 3 ˆ R 0 ˆ B 3 ψ (0) 0 aver = i ¯ he mc 2 AB jl γ A γ B ψ (0) 0 ∇ j · I A ×r Aj r 3 Aj ˆ R 0 ∇ l · I B ×r Bl r 3 Bl ψ (0) 0 aver = i ¯ he mc 2 AB jl γ A γ B ψ (0) 0 ∇ j · r Aj ×I A r 3 Aj ˆ R 0 ∇ l · r Bl ×I B r Bl ψ (0) 0 aver =− ¯ he mc 2 AB jl γ A γ B × ψ (0) 0 I A · ∇ j × r Aj r 3 Aj ˆ R 0 I B · ∇ l × r Bl r Bl ψ (0) 0 aver where the subscript “aver” means the averaging of eq. (W.4) and the identity A · (B ×C) = (A ×B) ·C has been used. We have the following chain of equalities . infinity and the operator is independent of the number of electrons in the system. The proof of the I Slater–Condon rule relies on the following chain of equalities ψ ˆ Gψ = 1 2 ij kl ij. sum of the one-electron operators 7 ˆ h(i) acting on functions of the coordinates of electron i. The I Slater–Condon rule says (see Appendix M), that for the Slater determi- nant ψ built of the. already have a differ- ence of one spinorbital. Thus, first we have to get rid of these spinorbitals (i and i ). For the integral to survive 9 we have to have at least one of the following conditions satisfied: •