10.15 EQUATION-OF-MOTION METHOD EOM-CC The CC method is used to calculate the ground state energy and wave function.. 10.15.1 SIMILARITY TRANSFORMATION Let us note that for the Schröding
Trang 1546 10 Correlation of the Electronic Motions
Taking advantage of the commutator expansion we have
mn
abe− ˆT 2Heˆ ˆT2
0
=
mn ab
1− ˆT2+1
2 ˆT2
2+ · · ·
ˆ H
1+ ˆT2+1
2 ˆT2
2+ · · ·
0
= mnab ˆH0
+ mnab ˆH ˆT20
+1 2
mn
ab ˆH ˆT220
− mnab ˆT2Hˆ 0
− mn
ab ˆT2H ˆˆT20 + A = 0
However,
A= −1
2
mn
ab ˆT2H ˆˆT2
20 +1 2
mn
ab ˆT22Hˆ 0 +1
2
mn
ab ˆT22H ˆˆT20 +1
4
mn
ab ˆT22H ˆˆT2
20
= 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, ( ˆT†
2)mnab|
de-notes a double deexcitation79of 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 ˆH0
+ mn
ab ˆH ˆT20
+1 2
mn
ab ˆH ˆT220
− mn
ab ˆT2Hˆ 0 − mn
ab ˆT2H ˆˆT20 = 0
After several days80 of algebraic manipulations, we get the equations for the t amplitudes (for each tabmnamplitude one equation):
(εm+ εn− εa− εb)tabmn
= mn|ab −
p>q
mn|pq tpq
ab −
γ>δ
cd|ab tmn
cd
c p
cn|bp tmp
ac − cm|bp tnp
ac − cn|ap tmp
bc + cm|ap tnp
bc
c>d p>q
cd|pq tabpqtcdmn− 2tabmptcdnq+ tabnqtcdmp
− 2tacmntbdpq+ tpq
actbdmn + 4tacmptbdnq+ tnq
actbdmp
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.81First,
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,
Trang 2the left-hand side the first approximation to tabmnis82tabmn∼ 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,83with 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 N6where N is a number of molecular orbitals (occupied and virtual84),
whereas the analogous cost of the CCSDT (singles, doubles, triples) method
re-quires N8 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 two85 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 show86 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 ˆH= ˆHA+ ˆHB
In such a situation the wave operator can be expressed as
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 (εm+ε n −ε a −ε b ) is close to 0 (quasidegeneracy of the vacuum
state with some other state), because then tmnab → ∞.
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 N4.
85 This can be generalized to many non-interacting subsystems.
86B Jeziorski, J Paldus, P Jankowski, Intern J Quantum Chem 56 (1995) 129.
Trang 3548 10 Correlation of the Electronic Motions
where ˆTA, ˆTB ˆTAB include the operators corresponding to spinorbitals from the subsystems A B and from the system AB, respectively Of course, in this situation
we have the following commutation condition:
ˆHA ˆTB
= ˆHB ˆTA
Then, owing to the commutator expansion in eq (10.42), we obtain:
e− ˆT ˆHA+ ˆHB
eˆT = e− ˆTAHˆAeˆTA+ e− ˆTBHˆBeˆTB+ O ˆTAB
where O( ˆTAB) denotes the linear and higher terms in ˆTAB 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
mixedO ˆTAB
0
which, due to the linear term in O( ˆTAB), is fulfilled by ˆTAB= 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
Let us note that for the Schrödinger equation ˆHψ= Eψ, we can perform an inter-esting sequence of transformations based on the wave operator eˆT:
e− ˆTHψˆ = Ee− ˆTψ
e− ˆTHeˆ ˆTe− ˆTψ= Ee− ˆTψ
We obtain the eigenvalue equation again, but for the similarity transformed
Hamiltonian
ˆ
H ¯ψ = E ¯ψ
where ˆH = e− ˆTHeˆ ˆT, ¯ψ= e− ˆTψ, and the energy E does not change at all after this
transformation This result will be very useful in a moment
Trang 410.15.2 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:
ψ0= exp ˆT
where 0is usually a Hartree–Fock determinant Now, we define the operator ˆUk
(“EOM-CC Ansatz”), which performs a miracle: from the wave function of the
ground state ψ0 it creates the wave function ψk for the k-th excited state of the
system:
ψk= ˆUkψ0 The operators ˆUk change the coefficients in front of the configurations (see
p 526) The operators ˆUk are (unlike the wave operator exp( ˆT )) linear with
re-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 ˆT in
the form of the sum of single and double excitations:
ˆT = ˆT1+ ˆT2
and
ˆUk= ˆUk 0+ ˆUk 1+ ˆUk 2 where the task for the ˆUk 0operator is to change the coefficient in front of the
func-tion 0to that appropriate to the|k function The role of the operators ˆUk 1 ˆUk 2
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 amam-plitudes of the CC method):
ˆUk 0= τ0(k)
ˆUk 1=
a p
τpa(k)ˆp†ˆa
ˆUk 2=
a b p q
τpqab(k)ˆq†ˆp†ˆa ˆb
where the amplitudes τ(k) are numbers, which are the targets of the EOM-CC
method The amplitudes give the wave function ψkand the energy Ek
We write down the Schrödinger equation for the excited state:
ˆ
Hψk= Ekψk
Now we substitute the EOM-CC Ansatz:
ˆ
H ˆUkψ0= EkˆUkψ0 and from the definition of the CC wave operator we get87
ˆ
H ˆUkexp ˆT
0= EkˆUkexp ˆT
0
87 By neglecting higher than single and double excitations the equation represents an approximation.
Trang 5550 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 ˆUkand ˆT commute,88hence the oper-ators ˆUkand exp( ˆT ) also commute:
ˆUkexp ˆT
= exp ˆT ˆUk Substituting this we have:
ˆ
H exp ˆT ˆUk0= Ekexp ˆT ˆUk0 and multiplying from the left with exp(− ˆT ) we get:
exp
− ˆT ˆH exp ˆT ˆUk0= Ek ˆUk0
or introducing the similarity transformed Hamiltonian
ˆ
H = e− ˆTHeˆ ˆT
we obtain
ˆ
H ˆUk0= EkˆUk0 From the last equation we subtract the CC equation for the ground state
exp
− ˆT ˆH exp ˆT
0= E00 multiplied from the left with ˆUk, i.e ˆUkHˆ 0= E0ˆUk0and we get
ˆ
H ˆUk0− ˆUkHˆ 0= EkˆUk0− E0ˆUk0 Finally, we obtain an important result:
ˆH ˆUk
0= (Ek− E0) ˆUk0 The operator ˆUkcontains 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 ˆUk, including89that 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 ψk
88 If ˆ Ukcontains true excitations, then it does not matter whether excitations are performed by ˆ UkˆT
or ˆ T ˆ Uk(commutation), because both ˆ Ukand ˆ T mean going up in the energy scale If, however, ˆ Uk contains deexcitations, then it may happen that there is an attempt in ˆ T ˆ Ukto deexcite the ground state wave function – that makes immediately 0, whereas ˆ UkˆT may be still OK, because the excitations in ˆT may be more important than the deexcitations in ˆ Uk.
89 More precisely: to get only the excitation energy we do not need the coefficient next to
Trang 6The unknown parameters are amplitudes and the excitation energies Ek− E0:
mn
ab[ ˆH ˆUk]0
= (Ek− E0)mnab ˆUk0
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 syssys-tem plus the first-order correction plus the
second-order correction plus , etc.90If the perturbation is small then we hope91
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 ψ0
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.92A Slater
deter-minant Iis 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
ˆF(i) =
i
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.
Trang 7552 10 Correlation of the Electronic Motions
Indeed,
ˆ
H(0)I=
i
εiˆı†ˆı · I=
i
εi
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 φiis present in ψ(0)0
The eigenvalue of ˆH0=iεiˆı†ˆı 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 ˆH(0)
10.16.2 PERTURBATION THEORY – SLIGHTLY DIFFERENT APPROACH
We have to solve the Schrödinger equation for the ground state93Hψˆ 0= Eψ0, with ˆ
H= ˆH(0)+ ˆH(1), where ˆH(0)denotes the unperturbed Hamiltonian, and ˆH(1)is a perturbation operator We assumed that ˆH(0)has eigenfunctions and correspond-ing energy eigenvalues
ˆ
H(0)ψ(0)k = E(0)
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
ψ0|ψ(0)
0 = 1 This means that the (non-normalized) ψ0must include the normal-ized function of zeroth order ψ(0)0 and, possibly, something orthogonal to it.
Let us write ˆHψ0as ˆHψ0= ( ˆH(0)+ ˆH(1))ψ0, or, in another way, as ˆH(1)ψ0= ( ˆH− ˆH(0))ψ0 Multiplying this equation by ψ(0)0 and integrating, we get (taking advantage of the intermediate normalization)
ψ(0)0 ˆH(1)ψ0
= ψ(0)0 ˆH− ˆH(0)
ψ0
= E0ψ(0)0 ψ0
− ψ(0)0 ˆH(0)ψ0
= E0− E(0)
Thus,
E0= ψ(0)0 ˆH(1)ψ0
93 We use the notation from Chapter 5.
Trang 8Fig 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
ˆP = |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
ˆP =ψ(0)
This means that ˆPχ is, within accuracy to a constant, equal to ψ(0)0 for an arbitrary
function χ Indeed, if χ 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)
n
then (Fig 10.11.b)
ˆPχ =
n
cn ˆPψ(0)
n =
n
cnψ(0)
0 ψ(0)0 ψ(0)
n
n
cnδ0nψ(0)0 = c0ψ(0)0 (10.62)
Let us now introduce a projection operator
ˆ
Q= 1 − ˆP =∞
n =1
ψ(0)
in the space orthogonal to ψ(0)0 Obviously, ˆP ˆQ= 0, ˆP2= ˆP and ˆQ2= ˆQ The latter
holds since ˆQ2= (1 − ˆP)2= 1 − 2 ˆP + ˆP2= 1 − ˆP = ˆQ
Trang 9554 10 Correlation of the Electronic Motions
Now we define a reduced resolvent
reduced
resolvent
ˆR0=∞
n =1
|ψ(0)
n ψ(0)
n |
E0(0)− E(0)
n
For functions orthogonal to ψ(0)0 , the action of the operator ˆR0 is identical to that of the operator (E0(0)− ˆH(0))−1 Let us make sure of this Let us operate first
on the function φ orthogonal to ψ(0)0 with the operator ˆR0(E0(0)− ˆH(0)) The result should be equal to φ Let us see:
ˆR0E0(0)− ˆH(0)
φ=∞
n =1
E0(0)− E(0)
n
−1ψ(0)
n ψ(0)n E(0)
0 − ˆH(0)φ
(10.65)
=
∞
n=1
E0(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
since for φ orthogonal to ψ(0)0 the projection ˆQφ equals φ Let us now operate
on the same function with the operator (E0(0)− ˆH(0)) ˆR0(i.e the operators are in reverse order):
E0(0)− ˆH(0) ˆR0φ=
E0(0)− ˆH(0)∞
n =1
E0(0)− E(0)
n
−1ψ(0)
n ψ(0)n φ
=
∞
n=1
E(0)0 − E(0)
n
−1
E0(0)− ˆH(0)ψ(0)
n ψ(0)n φ
=
∞
n =1
ψ(0)
n ψ(0)n φ
It really looks as if the ˆR0is the inverse of (E0(0)− ˆH(0)) This is not so, since when acting on the function ψ(0)0 we get
ˆR0E0(0)− ˆH(0)
and not ψ(0)0 In other words ˆR0(E0(0)− ˆH(0))ψ(0)0 = 0 = ψ(0)
0 Similarly,
E0(0)− ˆH(0) ˆR0ψ(0)0 =E0(0)− ˆH(0)∞
n=1
E0(0)− E(0)
n
−1ψ(0)
n ψ(0)n ψ(0)
0
=E0(0)− ˆH(0)
· 0 = 0 = ψ(0)
0
Trang 10Thus, the reduced resolvent is “almost” the inverse of (E0(0)− ˆH(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 ˆR0on function φ:
ˆR0φ=∞
n =1
E0(0)− E(0)
n
−1ψ(0)
n ψ(0)n φ
= linear combination of functions orthogonal to ψ(0)
Such a linear combination always belongs to the Q space, but we have not obtained
φ, hence ˆR0is 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
E0− ˆH(0)
We aim at having (E(0)0 − ˆH(0))ψ0on the left-hand side Let us add (E(0)0 − E0)ψ0
to both sides of that equation to obtain
E0− ˆH(0)
ψ0+E0(0)− E0
ψ0= ˆH(1)ψ0+E0(0)− E0
E(0)0 − ˆH(0)
ψ0=E(0)0 − E0+ ˆH(1)
Let us now operate on both sides of this equation with the reduced resolvent ˆR0
ˆR0E(0)0 − ˆH(0)
ψ0= ˆR0
E0(0)− E0+ ˆH(1)
On the left-hand side we have ˆQψ0(as follows from eq (10.67)), but ˆQψ0= (1 −
ˆP)ψ0= ψ0− |ψ(0)
0 ψ(0)
0 |ψ0 = ψ0− ψ(0)
0 , due to the intermediate normalization
As a result, the equation takes the form
ψ0− ψ(0)
0 = ˆR0
E(0)0 − E0+ ˆH(1)
Thus, we obtain
ψ0= ψ(0)
0 + ˆR0
E(0)0 − E0+ ˆH(1)
At the same time, based on the expression for E in perturbation theory
(eq (10.59)), we have:
E0= E(0)
0 + ψ(0)0 ˆH(1)ψ0
... amplitudes and the energy of the AB system is bound to be the sum of the energies of subsystemA and subsystem B (size consistency)
10.15 EQUATION -OF- MOTION METHOD (EOM-CC)...
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 ˆH(0)
10.16.2...
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