546 10. Correlation of the Electronic Motions Taking advantage of the commutator expansion we have mn ab e − ˆ T 2 ˆ He ˆ T 2 0 = mn ab 1 − ˆ T 2 + 1 2 ˆ T 2 2 +··· ˆ H 1 + ˆ T 2 + 1 2 ˆ T 2 2 +··· 0 = mn ab ˆ H 0 + mn ab ˆ H ˆ T 2 0 + 1 2 mn ab ˆ H ˆ T 2 2 0 − mn ab ˆ T 2 ˆ H 0 − mn ab ˆ T 2 ˆ H ˆ T 2 0 +A =0 However, A =− 1 2 mn ab ˆ T 2 ˆ H ˆ T 2 2 0 + 1 2 mn ab ˆ T 2 2 ˆ H 0 + 1 2 mn ab ˆ T 2 2 ˆ H ˆ T 2 0 + 1 4 mn ab ˆ T 2 2 ˆ H ˆ T 2 2 0 = 0 The last equality follows from the fact that each term is equal to zero. The first vanishes since both determinants differ by four excitations. Indeed, ( ˆ T † 2 ) mn ab | de- notes a double deexcitation 79 of the doubly excited function, i.e. something propor- excitations and deexcitations tional to 0 | For similar reasons (too strong deexcitations give zero) the remain- ing terms in A also vanish. As a result we need to solve the equation: mn ab ˆ H 0 + mn ab ˆ H ˆ T 2 0 + 1 2 mn ab ˆ H ˆ T 2 2 0 − mn ab ˆ T 2 ˆ H 0 − mn ab ˆ T 2 ˆ H ˆ T 2 0 =0 After several days 80 of algebraic manipulations, we get the equations for the t amplitudes (for each t mn ab amplitude one equation): (ε m +ε n −ε a −ε b )t mn ab =mn|ab− p>q mn|pqt pq ab − γ>δ cd|abt mn cd + cp cn|bpt mp ac −cm|bpt np ac −cn|apt mp bc +cm|apt np bc + c>d p>q cd|pq t pq ab t mn cd −2 t mp ab t nq cd +t nq ab t mp cd −2 t mn ac t pq bd +t pq ac t mn bd +4 t mp ac t nq bd +t nq ac t mp bd (10.49) It can be seen that the last expression includes: the term independent of t,the linear terms and the quadratic terms. How can we find the t’s? We do it with the help of the iterative method. 81 First, we substitute zeros for all t’s on the right-hand side of the equation. Thus, from 79 Opposite to excitation. 80 Students – more courage! 81 We organize things in such a way that a given unknown parameter will occur in the simple form on one side of the equation, whereas the more complicated terms, also containing the parameter sought, 10.14 Coupled cluster (CC) method 547 the left-hand side the first approximation to t mn ab is 82 t mn ab ∼ = mn|ab (ε m +ε n −ε a −ε b ) We have now an estimate of each amplitude – we are making progress. The approximation to t obtained in this way is substituted into the right-hand side to evaluate the left-hand side and so forth. Finally, we achieve a self-consistency of the iterative process and obtain the CC wave function for the ground state of our system. With the amplitudes we calculate the energy of the system with eq. (10.45). This is how the CCD (the CC with double excitations in the cluster operator) works from the practical viewpoint. It is more efficient when the initial amplitudes are taken from a short CI expansion, 83 with subsequent linearization (as above) of terms containing the initial (known) amplitudes. The computational cost of the CCD and CCSD (singles and doubles) methods scales as N 6 where N is a number of molecular orbitals (occupied and virtual 84 ), whereas the analogous cost of the CCSDT (singles, doubles, triples) method re- quires N 8 scaling. This means that, if we increase the orbital basis twice, the in- crease in the computational cost of the CCSDT method will be four times larger than that of the CCSD scheme. This is a lot, and because of this, wide-spread pop- ularity has been gained for the CCSD(T) method, which only partly uses the triple excitations. 10.14.5 SIZE CONSISTENCY OF THE CC METHOD The size consistency of the CC method can be proved on the basis of eq. (10.43) and (10.44). Let us assume that the system dissociates into two 85 non-interacting subsystems A and B (i.e. at infinite distance). Then the orbitals can be also divided into two separable (mutually orthogonal) subsets. We will show 86 that the cluster amplitudes, having mixed indices (from the first and second groups of orbitals), are equal to 0. Let us note first that, for an infinite distance, the Hamiltonian ˆ H = ˆ H A + ˆ H B . In such a situation the wave operator can be expressed as ˆ T = ˆ T A + ˆ T B + ˆ T AB (10.50) are kept on the other side of the equation. Then we take a certain random value of the unknown and calculate the complicated side of the equation. On the left-hand side we will then have its new approximation to the true value. We repeat the whole procedure so many times until we start getting the same value (if the procedure converges). Then the equation is solved. 82 As we see we would have trouble if (ε m +ε n −ε a −ε b ) is close to 0 (quasidegeneracy of the vacuum state with some other state), because then t mn ab →∞. 83 The configuration interaction method with inclusion of single and double excitations only: CCD: J.A. Pople, R. Krishnan, H.B. Schlegel, J.S. Binkley, Intern. J. Quantum Chem. S14 (1978) 545; R.J. Bartlett, G.D. Purvis III, Intern. J. Quantum Chem. S14 (1978) 561. CCSD: G.D. Purvis III, J. Chem. Phys. 76 (1982) 1910. 84 These estimations are valid for the same relative increase of the number of occupied and virtual orbitals, as it is, e.g., for going from a molecule to its dimer. In the case of calculations for the same molecule, but two atomic basis sets (that differ in size) the cost increases only as N 4 . 85 This can be generalized to many non-interacting subsystems. 86 B. Jeziorski, J. Paldus, P. Jankowski, Intern. J. Quantum Chem. 56 (1995) 129. 548 10. Correlation of the Electronic Motions where ˆ T A , ˆ T B ˆ T AB include the operators corresponding to spinorbitals from the subsystems AB and from the system AB, respectively. Of course, in this situation we have the following commutation condition: ˆ H A ˆ T B = ˆ H B ˆ T A =0 (10.51) Then, owing to the commutator expansion in eq. (10.42), we obtain: e − ˆ T ˆ H A + ˆ H B e ˆ T =e − ˆ T A ˆ H A e ˆ T A +e − ˆ T B ˆ H B e ˆ T B +O ˆ T AB (10.52) where O( ˆ T AB ) denotes the linear and higher terms in ˆ T AB Substituting this into eq. 10.44 with bra mixed| vector representing mixed excitation, we observe that the first two terms on the right-hand side of the last equation give zero. It means that we get the equation mixed O ˆ T AB 0 =0 (10.53) which, due to the linear term in O( ˆ T AB ), is fulfilled by ˆ T AB = 0 Conclusion: for the infinite distance between the subsystems we do not have mixed amplitudes and the energy of the AB system is bound to be the sum of the energies of subsystem A and subsystem B (size consistency). 10.15 EQUATION-OF-MOTION METHOD (EOM-CC) The CC method is used to calculate the ground state energy and wave function. What about the excited states? This is a task for the equation-of-motion CC method, the primary goal being not the excited states themselves, but the exci- tation energies with respect to the ground state. 10.15.1 SIMILARITY TRANSFORMATION LetusnotethatfortheSchrödingerequation ˆ Hψ =Eψ, we can perform an inter- esting sequence of transformations based on the wave operator e ˆ T : e − ˆ T ˆ Hψ =Ee − ˆ T ψ e − ˆ T ˆ He ˆ T e − ˆ T ψ = Ee − ˆ T ψ We obtain the eigenvalue equation again, but for the similarity transformed Hamiltonian ˆ H ¯ ψ =E ¯ ψ where ˆ H =e − ˆ T ˆ He ˆ T , ¯ ψ =e − ˆ T ψ, and the energy E does not change at all after this transformation. This result will be very useful in a moment. 10.15 Equation-of-motion method (EOM-CC) 549 10.15.2 DERIVATION OF THE EOM-CC EQUATIONS As the reference function in the EOM-CC method, we take the coupled-cluster wave function for the ground state: ψ 0 =exp ˆ T 0 (10.54) where 0 is usually a Hartree–Fock determinant. Now, we define the operator ˆ U k (“EOM-CC Ansatz”), which performs a miracle: from the wave function of the ground state ψ 0 it creates the wave function ψ k for the k-th excited state of the system: ψ k = ˆ U k ψ 0 The operators ˆ U k change the coefficients in front of the configurations (see p. 526). The operators ˆ U k are (unlike the wave operator exp( ˆ T))linearwithre- spect to the excitations, i.e. the excitation amplitudes occur there in the first pow- ers. For the case of the single and double excitations (EOM-CCSD) we have ˆ T in the form of the sum of single and double excitations: ˆ T = ˆ T 1 + ˆ T 2 and ˆ U k = ˆ U k0 + ˆ U k1 + ˆ U k2 where the task for the ˆ U k0 operator is to change the coefficient in front of the func- tion 0 to that appropriate to the |kfunction. The role of the operators ˆ U k1 ˆ U k2 is an appropriate modification of the coefficients in front of the singly and doubly excited configurations. These tasks are done by the excitation operators with τ am- plitudes (they have to be distinguished from the t amplitudes of the CC method): ˆ U k0 = τ 0 (k) ˆ U k1 = ap τ p a (k) ˆ p † ˆ a ˆ U k2 = abpq τ pq ab (k) ˆ q † ˆ p † ˆ a ˆ b where the amplitudes τ(k) are numbers, which are the targets of the EOM-CC method. The amplitudes give the wave function ψ k and the energy E k . We write down the Schrödinger equation for the excited state: ˆ Hψ k =E k ψ k Now we substitute the EOM-CC Ansatz: ˆ H ˆ U k ψ 0 =E k ˆ U k ψ 0 and from the definition of the CC wave operator we get 87 ˆ H ˆ U k exp ˆ T 0 =E k ˆ U k exp ˆ T 0 87 By neglecting higher than single and double excitations the equation represents an approximation. 550 10. Correlation of the Electronic Motions Due to the missing deexcitation part (i.e. that which lowers the excitation rank, e.g., from doubles to singles) the operators ˆ U k and ˆ T commute, 88 hence the oper- ators ˆ U k and exp( ˆ T)also commute: ˆ U k exp ˆ T =exp ˆ T ˆ U k Substituting this we have: ˆ H exp ˆ T ˆ U k 0 =E k exp ˆ T ˆ U k 0 and multiplying from the left with exp(− ˆ T)we get: exp − ˆ T ˆ H exp ˆ T ˆ U k 0 =E k ˆ U k 0 or introducing the similarity transformed Hamiltonian ˆ H =e − ˆ T ˆ He ˆ T we obtain ˆ H ˆ U k 0 =E k ˆ U k 0 From the last equation we subtract the CC equation for the ground state exp − ˆ T ˆ H exp ˆ T 0 =E 0 0 multiplied from the left with ˆ U k , i.e. ˆ U k ˆ H 0 =E 0 ˆ U k 0 and we get ˆ H ˆ U k 0 − ˆ U k ˆ H 0 =E k ˆ U k 0 −E 0 ˆ U k 0 Finally, we obtain an important result: ˆ H ˆ U k 0 =(E k −E 0 ) ˆ U k 0 The operator ˆ U k contains the sought amplitudes τ(k). We find them in a similar manner as in the CC method. For that purpose we make a scalar product of the left- and right-hand side of that equation with each excitation | mn ab used in ˆ U k , including 89 that of no excitation, i.e. the function 0 . We get the set of the EOM-CC equations whose number is equal to the number of sought amplitudes plus one more equation due to normalization condition of ψ k . 88 If ˆ U k contains true excitations, then it does not matter whether excitations are performed by ˆ U k ˆ T or ˆ T ˆ U k (commutation), because both ˆ U k and ˆ T mean going up in the energy scale. If, however, ˆ U k contains deexcitations, then it may happen that there is an attempt in ˆ T ˆ U k to deexcite the ground state wave function – that makes immediately 0, whereas ˆ U k ˆ T may be still OK, because the excitations in ˆ T may be more important than the deexcitations in ˆ U k . 89 More precisely: to get only the excitation energy we do not need the coefficient next to 0 . 10.16 Many body perturbation theory (MBPT) 551 The unknown parameters are amplitudes and the excitation energies E k −E 0 : mn ab [ ˆ H ˆ U k ] 0 =(E k −E 0 ) mn ab ˆ U k 0 We solve these equations and the problem is over. 10.16 MANY BODY PERTURBATION THEORY (MBPT) The majority of routine calculations in quantum chemistry are done with varia- tional methods (mainly the Hartree–Fock scheme). If we consider post-Hartree– Fock calculations then non-variational [CCSD, CCSD(T)] as well as perturbational (among them MBPT) approaches take the lead. The perturbational methods are based on the simple idea that the system, in a slightly modified condition, is similar to that before the perturbation is applied (cf. p. 203). In the formalism of perturbation theory, knowing the unperturbed system and the perturbation we are able to provide successive corrections to obtain the solu- tion of the perturbed system. Thus, for instance, the energy of the perturbed sys- tem is the energy of the unperturbed system plus the first-order correction plus the second-order correction plus,etc. 90 If the perturbation is small then we hope 91 the series is convergent, even then however, there is no guarantee that the series converges fast. 10.16.1 UNPERTURBED HAMILTONIAN In the perturbational approach (cf. p. 204) to the electron correlation the Hartree–Fock function, 0 , is treated as the zero-order approximation to the true ground state wave function, i.e. 0 =ψ (0) 0 . Thus, the Hartree–Fock wave function stands at the starting point, while the goal is the exact ground- state electronic wave function ψ 0 . In majority of cases this is a reasonable approximation, since the Hartree–Fock method usually provides as much as 98–99% of the total energy. 92 A Slater deter- minant I is constructed from the spinorbitals satisfying the Fock equation. How to construct the operator for which the Slater determinant is an eigenfunction? We will find out in a moment that this operator is the sum of the Fock operators (cf. Appendix U) ˆ H (0) = i ˆ F(i)= i ε i ˆ ı † ˆ ı (10.55) 90 This is an old trick of perturbation theory, equivalent to saying that the shape of a bridge loaded with a car is the shape of the bridge without the car plus the deformation proportional to the mass of the car plus the deformation proportional to the square of the mass of the car, etc. This works, if the bridge is solid and the car is light (the perturbation is small). 91 There is not much known concerning the convergence of series occurring in quantum chemistry. Commonly, only a few perturbational corrections are computed. 92 Sometimes, as we know, the method fails and then the perturbation theory based on the Hartree– Fock starting point is a risky business, since the perturbation is very large. 552 10. Correlation of the Electronic Motions Indeed, ˆ H (0) I = i ε i ˆ ı † ˆ ı · I = i ε i · I (10.56) since the annihilation of one spinorbital in the determinant and the creation of the same spinorbital leaves the determinant unchanged. This is so on condition that the spinorbital φ i is present in ψ (0) 0 . The eigenvalue of ˆ H 0 = i ε i ˆ ı † ˆ ı is always the sum of the orbital energies corresponding to all spinorbitals in the Slater determinant I . This means that the sum of several determinants, each built from a different (in the sense of the orbital energies) set of spinorbitals, is not an eigenfunction of ˆ H (0) . 10.16.2 PERTURBATION THEORY – SLIGHTLY DIFFERENT APPROACH We have to solve the Schrödinger equation for the ground state 93 ˆ Hψ 0 =Eψ 0 ,with ˆ H = ˆ H (0) + ˆ H (1) ,where ˆ H (0) denotes the unperturbed Hamiltonian, and ˆ H (1) is a perturbation operator. We assumed that ˆ H (0) has eigenfunctions and correspond- ing energy eigenvalues ˆ H (0) ψ (0) k =E (0) k ψ (0) k (10.57) The ground state ψ (0) 0 is non-degenerate (assumption). The Schrödinger equation does not force the normalization of the function. It is convenient to use the intermediate normalization (Fig. 10.11.a), i.e. to require that ψ 0 |ψ (0) 0 =1. This means that the (non-normalized) ψ 0 must include the normal- ized function of zeroth order ψ (0) 0 and, possibly, something orthogonal to it. Let us write ˆ Hψ 0 as ˆ Hψ 0 =( ˆ H (0) + ˆ H (1) )ψ 0 , or, in another way, as ˆ H (1) ψ 0 = ( ˆ H − ˆ H (0) )ψ 0 . Multiplying this equation by ψ (0) 0 and integrating, we get (taking advantage of the intermediate normalization) ψ (0) 0 ˆ H (1) ψ 0 = ψ (0) 0 ˆ H − ˆ H (0) ψ 0 =E 0 ψ (0) 0 ψ 0 − ψ (0) 0 ˆ H (0) ψ 0 = E 0 −E (0) 0 =E 0 (10.58) Thus, E 0 = ψ (0) 0 ˆ H (1) ψ 0 (10.59) 93 We use the notation from Chapter 5. 10.16 Many body perturbation theory (MBPT) 553 projection Fig. 10.11. Pictorial presentation of (a) the intermediate normalization ψ|ψ (0) 0 =1, ψ (n) 0 is the n-th correction, and (b) the projection onto the axis ψ (0) 0 in the Hilbert space using the operator ˆ P =|ψ (0) 0 ψ (0) 0 |. 10.16.3 REDUCED RESOLVENT OR THE “ALMOST” INVERSE OF (E (0) 0 − ˆ H (0) ) Let us define several useful quantities – we need to get familiar with them now – which will introduce a certain elegance into our final equations. Let the first be a projection operator on the ground-state zeroth order function projection operator ˆ P = ψ (0) 0 ψ (0) 0 (10.60) This means that ˆ Pχ is, within accuracy to a constant, equal to ψ (0) 0 for an arbitrary function χ.Indeed,ifχ is expressed as a linear combination of the eigenfunctions ψ (0) n (these functions form an orthonormal complete set as eigenfunctions of the Hermitian operator) χ = n c n ψ (0) n (10.61) then (Fig. 10.11.b) ˆ Pχ = n c n ˆ Pψ (0) n = n c n ψ (0) 0 ψ (0) 0 ψ (0) n = n c n δ 0n ψ (0) 0 =c 0 ψ (0) 0 (10.62) Let us now introduce a projection operator ˆ Q =1 − ˆ P = ∞ n=1 ψ (0) n ψ (0) n (10.63) in the space orthogonal to ψ (0) 0 . Obviously, ˆ P ˆ Q =0, ˆ P 2 = ˆ P and ˆ Q 2 = ˆ Q. The latter holds since ˆ Q 2 =(1 − ˆ P) 2 =1 −2 ˆ P + ˆ P 2 =1 − ˆ P = ˆ Q. 554 10. Correlation of the Electronic Motions Now we define a reduced resolvent reduced resolvent ˆ R 0 = ∞ n=1 |ψ (0) n ψ (0) n | E (0) 0 −E (0) n (10.64) For functions orthogonal to ψ (0) 0 , the action of the operator ˆ R 0 is identical to that of the operator (E (0) 0 − ˆ H (0) ) −1 . Let us make sure of this. Let us operate first on the function φ orthogonal to ψ (0) 0 with the operator ˆ R 0 (E (0) 0 − ˆ H (0) ). The result should be equal to φ. Let us see: ˆ R 0 E (0) 0 − ˆ H (0) φ = ∞ n=1 E (0) 0 −E (0) n −1 ψ (0) n ψ (0) n E (0) 0 − ˆ H (0) φ (10.65) = ∞ n=1 E (0) 0 −E (0) n −1 E (0) 0 −E (0) n ψ (0) n ψ (0) n φ (10.66) = ∞ n=1 ψ (0) n ψ (0) n φ = ˆ Qφ =φ (10.67) since for φ orthogonal to ψ (0) 0 the projection ˆ Qφ equals φ.Letusnowoperate on the same function with the operator (E (0) 0 − ˆ H (0) ) ˆ R 0 (i.e. the operators are in reverse order): E (0) 0 − ˆ H (0) ˆ R 0 φ = E (0) 0 − ˆ H (0) ∞ n=1 E (0) 0 −E (0) n −1 ψ (0) n ψ (0) n φ = ∞ n=1 E (0) 0 −E (0) n −1 E (0) 0 − ˆ H (0) ψ (0) n ψ (0) n φ = ∞ n=1 ψ (0) n ψ (0) n φ = ˆ Qφ =φ (10.68) It really looks as if the ˆ R 0 is the inverse of (E (0) 0 − ˆ H (0) ).Thisisnotso,since when acting on the function ψ (0) 0 we get ˆ R 0 E (0) 0 − ˆ H (0) ψ (0) 0 = ˆ R 0 ·0 =0 (10.69) and not ψ (0) 0 . In other words ˆ R 0 (E (0) 0 − ˆ H (0) )ψ (0) 0 =0 =ψ (0) 0 Similarly, E (0) 0 − ˆ H (0) ˆ R 0 ψ (0) 0 = E (0) 0 − ˆ H (0) ∞ n=1 E (0) 0 −E (0) n −1 ψ (0) n ψ (0) n ψ (0) 0 = E (0) 0 − ˆ H (0) ·0 =0 =ψ (0) 0 10.16 Many body perturbation theory (MBPT) 555 Thus, the reduced resolvent is “almost” the inverse of (E (0) 0 − ˆ H (0) ), almost, be- cause it happens only when acting on the functions from the space orthogonal to ψ (0) 0 . When the reduced resolvent operates on an arbitrary function, the result be- longs to the Q space, but it does not represent a projection on the Q space. Indeed, let us operate with ˆ R 0 on function φ: ˆ R 0 φ = ∞ n=1 E (0) 0 −E (0) n −1 ψ (0) n ψ (0) n φ = linear combination of functions orthogonal to ψ (0) 0 (10.70) Such a linear combination always belongs to the Q space, but we have not obtained φ,hence ˆ R 0 is not a projection operator. 10.16.4 MBPT MACHINERY Our goal now will be to present the Schrödinger equation in a different form. Let us first write it down as follows E 0 − ˆ H (0) ψ 0 = ˆ H (1) ψ 0 (10.71) We aim at having (E (0) 0 − ˆ H (0) )ψ 0 on the left-hand side. Let us add (E (0) 0 −E 0 )ψ 0 to both sides of that equation to obtain E 0 − ˆ H (0) ψ 0 + E (0) 0 −E 0 ψ 0 = ˆ H (1) ψ 0 + E (0) 0 −E 0 ψ 0 (10.72) or E (0) 0 − ˆ H (0) ψ 0 = E (0) 0 −E 0 + ˆ H (1) ψ 0 (10.73) Let us now operate on both sides of this equation with the reduced resolvent ˆ R 0 ˆ R 0 E (0) 0 − ˆ H (0) ψ 0 = ˆ R 0 E (0) 0 −E 0 + ˆ H (1) ψ 0 (10.74) On the left-hand side we have ˆ Qψ 0 (as follows from eq. (10.67)), but ˆ Qψ 0 =(1 − ˆ P)ψ 0 =ψ 0 −|ψ (0) 0 ψ (0) 0 |ψ 0 =ψ 0 −ψ (0) 0 , due to the intermediate normalization. As a result, the equation takes the form ψ 0 −ψ (0) 0 = ˆ R 0 E (0) 0 −E 0 + ˆ H (1) ψ 0 (10.75) Thus, we obtain ψ 0 =ψ (0) 0 + ˆ R 0 E (0) 0 −E 0 + ˆ H (1) ψ 0 (10.76) At the same time, based on the expression for E in perturbation theory (eq. (10.59)), we have: E 0 =E (0) 0 + ψ (0) 0 ˆ H (1) ψ 0 (10.77) . have mixed amplitudes and the energy of the AB system is bound to be the sum of the energies of subsystem A and subsystem B (size consistency). 10.15 EQUATION -OF- MOTION METHOD (EOM-CC) The CC method. coefficient in front of the func- tion 0 to that appropriate to the |kfunction. The role of the operators ˆ U k1 ˆ U k2 is an appropriate modification of the coefficients in front of the singly. product of the left- and right-hand side of that equation with each excitation | mn ab used in ˆ U k , including 89 that of no excitation, i.e. the function 0 . We get the set of the