Kapitel4 Cc Deriv

8/3/2019 Kapitel4 Cc Deriv

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Coupled cluster theory: Analytic derivatives,

molecular properties, and response theory

Wim Klopper and David P. Tew

Lehrstuhl fr Theoretische ChemieInstitut fr Physikalische Chemie

Universitt Karlsruhe (TH)

C4 Tutorial, Zrich, 24 October 2006

Variational and non-variational wavefunctions

A wavefunction is referred to as variational if the electronic

energy function E(x,) fulfills the condition

E(x,)

= 0 for all x

x is the molecular geometry or any other perturbationalparameter and represents the molecular electronic

wavefuntion parameters (e.g., MO coefficients, CC

amplitudes).

The electronic gradient vanishes at all geometries.

The variational condition determines as a function of x.

The molecular electronic energy is obtained by insertingthe optimal into the energy function.


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Examples of variational wavefunctions

The HartreeFock wavefunction is variational since the orbitalrotation parameters (MO coefficients) are variational,

EHF(x,)

= 0 for all x

The MCSCF wavefunction is variational since the variationalcondition is fulfilled both for the orbital rotation parameters (MO

coefficients) and the state transfer parameters p (CI coefficients),

EMCSCF(x,, p)

= 0,

EMCSCF(x,, p)

p= 0 for all x

CI wavefunctions are not variational since the variationalcondition is not fulfilled for the orbital rotation parameters ,

ECI(x,, p)

= 0,

ECI(x,, p)

p= 0 for all x

Derivatives of variational wavefunctions

The molecular electronic energy is obtained by inserting

the optimal electronic wavefunction parameters () into

the energy function,

(x) = E(x,)

We are interested in the first derivative

d(x)

dx=

E(x,)

x+

E(x,)

x

IfE(x,)

= 0, then

d(x)

dx=

E(x,)

x

We do not need the response of the variational

wavefunction!


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HellmannFeynman theorem

Assume that the (variational) energy function can be

written as an expectation value,

E(x,) = |H(x)|

We then obtain

d(x)

dx=

H(x)

x

This is the HellmannFeynman theorem.

Although originally stated for geometrical distortions,

it holds for any perturbation.

HellmannFeynman theorem for HartreeFock

Consider the Hamiltonian of a molecule in a static electric

field E,H(E) = H(0) E

Thus, in HartreeFock theory, the z-component of themolecules dipole can be computed as

d(x)

dEz=

H(E)Ez

= |z|

Concerning HartreeFock calculations in a finite basis,

note that the HellmannFeynman theorem does not hold

for a geometrical distortion (Ax),

d(x)

dAx=H(x)Ax


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HellmannFeynman force and corrections

In HartreeFock theory, a geometrical distortion (Ax) yieldsthe HellmannFeynman force plus corrections,

d(x)

dAx=

H(x)Ax

+ corrections

= ZA

Ni=1

xi Axr3A+ . . .

The reason is that the parameters of the one-electron

basis (exponents and contraction coefficients) arenon-variational electronic wavefunction parameters.

The corrections are sometimes called Pulay terms.

Second derivative of variational wavefunctions

The variational condition also simplifies the calculation of

second derivatives,

d2(x)

dx2 =

2E

x2 + 2

2E

x x + 2E2 x2

The term E/(2/x2) is eliminated by the variationalcondition.

We need the first-order response (/x) of the wavefunction to calculate the energy to second order.

2n+1 rule: The derivatives of the wavefunction to order n

determine the derivatives of the energy to order 2n+1.


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Response equations

The calculation of second derivatives requires the

knowledge of the first-order response /x.

This first-order response is obtained by differentiating the

equations that determine the electronic wavefunction

parameters .

For variational wavefunctions, the variational condition

E/ = 0 determines the parameters . Thus,

d

dx E = 2E

x +

2E

2 x = 0 We obtain a set of linear equations (response equations)

from which the first-order response may be determined.

Derivatives of non-variational wavefunctions

As an example, we consider the gradient of the CI energy,

which is variational w.r.t. the configuration coefficients p,but not w.r.t. the orbital rotations ,

ECI(x,, p)

p = 0,ECI(x,, p)

= 0

Therefore, if we differentiate the CI energy function w.r.t. x,we do not obtain the simplifications of the 2n+1 rule,

dCI(x)

dx=

ECI(x,, p)

x+

ECI(x,, p)

x

It appears that we need the first-order response of the

orbitals, /x.


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First-order response of the orbitals

The orbital rotation parameters are determined by the

variational HartreeFock condition

EHF(x,)

= 0 for all x

Thus, the first-order response of the orbitals /x can bedetermined by differentiating the HartreeFock condition

with respect to x,

2EHF(x,)

2

x = 2EHF(x,)

x

There is one set of response equations for each

perturbation, that is, for each independent geometrical

distortion.

Lagranges method of undetermined multipliers

By regarding the variational HartreeFock condition as a

set of constraints in the optimization of the CI energy, we

introduce the Lagrangian function

LCI(x,, , p) = ECI(x,, p) + EHF(x,)

are the Lagrange multipliers. The form of LCI is differentfrom ECI, but it gives the same energy when theHartreeFock condition is fulfilled.

We adjust the multipliers so that LCI becomes variationalin all variables. The price we pay for this is that there is a

larger number of variables.


7/31

The variational Lagrangian

The Lagrangian function is variational in all variables,

LCI(x,, , p)

p =ECI(x,, p)

p = 0

LCI(x,, , p)

=

EHF(x,)

= 0

LCI(x,, , p)

=

ECI(x,, p)

+

2EHF(x,)

2= 0

The last equation determines the Lagrange multipliers in

such a way that the Lagrangian is variational in .

With the Lagrangian function, we have a completely

variational formulation of the CI energy, and the total

derivative of the Lagrangian w.r.t. x is simply thecorresponding partial derivative.

The total derivative of the Lagrangian

The total derivative of the CI energy can be computed from

the Lagrangian,

dCI(x)

dx=

dLCI(x,, , p)

dx=

LCI(x,, , p)

x

=ECI(x,, p)

x+

2EHF(x,)x

The multipliers are obtained from the equation

2EHF(x,)

2 =

ECI(x,, p)

which does not depend on the perturbation x.

The perturbation independent formulation is also known as

Z-vector or interchange method.


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The coupled-cluster Lagrangian

The coupled-cluster energy is neither variational in the

orbital rotations nor in the amplitudes t

. Thus, we must

introduce Langrange multipliers and t,

LCC(x,, , t, t) = ECC(x,, t)+EHF(x,)

+t(x,, t)

where (x,, t) is the coupled-cluster vector function

(x,, t) = |HT(x)|HF = | exp(T)H(x) exp(T)|HF

The coupled-cluster amplitudes equations are

(x,, t) = 0 for all

Orbital-unrelaxed coupled-cluster properties

Let us first consider orbital-unrelaxed molecular properties.

Imagine that the perturbation is switched on only after the

HartreeFock calculation. Thus, the orbitals are not

changed by the perturbation and it suffices to consider the

unrelaxed Lagrangian

LCC,unrelaxed (x,, t, t) = ECC(x,, t) + t(x,, t)

The property can be obtained from

dLCC,unrelaxeddx

=LCC,unrelaxed

x=

ECCx

+ t

x

(Here and in the following we omit arguments for clarity.)


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A simple unrelaxed one-electron property

Consider (again) the Hamiltonian of a molecule in a static

electric field E,

H(E) = H(0) E

The (orbital-unrelaxed) z-component of the moleculesdipole can be computed as

z =ECCEz

+ t

Ez

Note that the z-component of the molecules dipole canalso be computed by means of finite perturbation theoryby adding the operator zEz after the HartreeFockcalculation has finished and before the coupled-cluster

calculation has begun.

The coupled-cluster multipliers

The coupled-cluster multipliers are obtained by requiring

that the coupled-cluster Lagrangian is variational in the

amplitudes,

LCC,unrelaxedt =

ECCt + t

t =

ECCt + t

= 0

where is the coupled-cluster Jacobian,

= | exp(T) [H(x), ] exp(T)|HF

Furthermore,

ECCt = HF|H(x)

|CC


10/31

The coupled-cluster HellmannFeynman theorem

Consider the following partial derivatives:

ECCx = HF H(x)x CC

t

x= t

exp(T)H(x)xCC

Thus, if we define a bra state

| = HF| + t| exp(T)

we can write the total derivative of the Lagrangian as

dLCC,unrelaxeddx

=

H(x)xCC

A variational coupled-cluster energy

The usual expression for the coupled-cluster energy is

(now omitting the x-dependence of H(x))

ECC = HF|H|CC = HF|HT|HF

Alternatively, we may compute the energy from

ECC,var = |H|CC = HF|HT|HF + t|H

T|HF

HF| + t| is the left eigenvector and |HF is the righteigenvector of the similarity-transformed Hamiltonian HT.

Of course, ECC,var is nothing but the CC Lagrangian.

The CC energy is less sensitive to numerical errors in theamplitudes and multipliers when evaluated from ECC,var.


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Coupled-cluster density matrices

Recall that the Hamiltonian in second quantization is

H = hnuc +PQ

hPQa

PaQ +12 PQRS

gPQRSa

Pa

RaSaQ

Hence, the energy ECC = HF|H|CC can be written as

ECC =PQ

DPQhPQ +12

PQRS

dPQRSgPQRS

DPQ = HF|aPaQ|CC, dPQRS = HF|a

Pa

RaSaQ|CC

The coupled-cluster density matrices are not Hermitian

and may give complex eigenvalues upon diagonalization.

For the energy, it is sufficient to consider the real

symmetric part.

Coupled-cluster Lagrangian density matrices

The energy ECC,var = |H|CC can be written as

ECC =

PQDPQhPQ +

12

PQRSdPQRSgPQRS

DPQ = |aPaQ|CC, d

PQRS = |aPa

RaSaQ|CC

In terms of the Lagrangian densities, we may calculate

coupled-cluster first-order properties in the same way as

for variational wavefunctions, contracting the density matrix

elements with the molecular integrals.

The Lagrangian density matrices are also known as the

variational or relaxed density matrices.


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A biorthogonal basis

We introduce the notation

(| = | exp(T) = HF| exp(T)

|) = exp(T)| = exp(T)|HF

These states form a biorthogonal set,

(|) =

For convenience, we identify 0 as the identity operator,

(0| = HF|0 exp(T) = HF| exp(T) = HF| = (HF|

|0) = exp(T)0|HF = exp(T)|HF = |CC = |HF)

Matrix representation of the Hamiltonian

We consider the matrix representation of the molecular

electronic Hamiltonian H in the biorthogonal basis,

H

= H00 H0H0 H with , > 0 H is an unsymmetric real matrix.

It follows that

H00 = (0|H|0) = HF| exp(T)Hexp(T)|HF = ECC

H0 = (|H|0) = | exp(T)Hexp(T)|HF = = 0

H0 = (0|H|) = HF| exp(T)Hexp(T)| = ECC/t


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Left and right eigenvectors

Apparently,H

= ECC ECC/t0 H with , > 0 ECC is an eigenvalue of H with right eigenvector

10

.

The left eigenvector (1 t) must fulfill

ECCt

+ t(H ECC) = 0

Recall the multipliers equation,

ECCt

+ t = 0 = H ECC

The coupled-cluster Jacobian

Earlier, we have encountered the Jacobian

=

t= | exp(T) [H, ] exp(T)|HF

= (| [H, ] |HF) = (|H|) (|H|HF)

= H (|H|HF)

We invoke the resolution of the identity to show that

(| H|HF) = |HT|HF =

|||HT|HF

= ||HFHF|HT|HF = ECC

Thus, the CC Jacobian occurs in the matrix representationof the similarity-transformed Hamiltonian.


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The EOM-CC eigenvalue problem

Differentiating the EOM-CC pseudo-expectation value w.r.t.

the ket and bra coefficients (assumed to be real), yields

Hck = Ekck

cTkH = c

Tk Ek

Note that

(ci|cj) = ci|cj = cTi cj = ij

(ci|H|cj) = ci| exp(T)Hexp(T)|cj

The EOM-CC states are obtained by diagonalizing the

unsymmetric matrix representation of the similarity-

transformed Hamiltonian. The ground-state amplitudes

are used in the similarity transformation.

Eigenvalues of the Jacobian

We shall level-shift the similarity-transformed Hamiltonianby the ground-state energy E0. The eigenvalues will thencorrespond to the excitation energies,

H = H E01 = 0 T0 , = E0t = HF|H |CCThus,

0 T

0

sktk

=

Ttktk

= Ek

sktk

The EOM-CC excitation energies correspond to the

eigenvalues of the CC Jacobian . Since is

unsymmetric, there is no guarantee that the eigenvalues

are real, but this is not a problem in practice.


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Some remarks on EOM-CC

The EOM-CC states are eigenvectors of the

similarity-transformed Hamiltonian (using ground-state

amplitudes). The excitation energies are eigenvalues ofthe ground-state CC Jacobian.

EOM-CC can be applied to the standard models CCSD,

CCSDT, etc.

An EOM-CC calculation on two non-interacting systems A

and B will recover the excitation energies of A and B

(size-intensivity), but simultaneous excitations in A and B

are not size-intensive.

For CCSD, CCSDT, etc., the EOM-CC excitation energies

are equal to those obtained from CC response theory.

Molecular gradients

So far, we have only considered orbital-unrelaxed molecularproperties. CC first-order properties can easily be computed

from the pseudo-expectation value |V|CC, that is, from thecorresponding variational density.

Next, consider a perturbation that changes the MOs (but not theAOs). The orbital-relaxed approach is now required. Consider,for example, a static electric field that is switched on already in

the HartreeFock step.

Matters become even more complicated when also the AO basisis perturbed. This happens, for example, when derivatives aretaken w.r.t. nuclear coordinates (molecular gradients), when the

metric is changed by relativistic perturbations, or when GIAOs

(London orbitals) are used for calculations of magneticproperties.


17/31

HartreeFock orbitals

The MOs are expanded in a basis of AOs,

P = cP

Thus, the derivative w.r.t. a nuclear coordinate becomes

P

x=

cPx

u + cP

x

Changes occur in the MO coefficients and in the AOs. Theproblem can be handled in a two-step procedure. At each

geometry x, we write the orthonormal HartreeFock orbitals as

CHF(x) = COMO(x)U(x)

where U(x) is a unitary (or orthogonal) matrix and COMO(x) abasis of orthonormal molecular orbitals (OMOs).

OMOs and UMOs

At the reference geometry x0, we choose U(x0) = 1 andCOMO(x0) = CHF(x0).

If the geometry changes from x0 to x, the unmodified molecularorbitals (UMOs) are no longer orthonormal,

CUMO(x) = COMO(x0)

S(x) = CTUMO(x)SAO(x)CUMO(x) = 1

We define orthonormalized molecular orbitals (OMOs),

COMO(x) = CUMO(x)S1/2(x)

Of course,

S(x) = CTOMO(x)SAO(x)COMO(x) = S1/2(x)S(x)S1/2(x) = 1


18/31

An orthogonal orbital connection

The connection matrixS1/2(x) connects orthonormalorbitals at neighbouring geometries. Rules that accomplish

this are called orbital connections. We use the OMOs (not the HartreeFock orbitals) to define

a Fock space, in which we represent the Hamiltonian insecond quantization,

H(x) = hnuc(x) +PQ

hPQ(x)aPaQ +

12

PQRS

gPQRS(x)aPa

RaS aQ

with

aP =Q

aQS1/2(x)

QP

aP =Q

aQ

S1/2(x)

PQ

Second quantization

We may ignore the geometry dependence of the creationand annihilation operators!

H(x) = hnuc(x) + PQhPQ(x)a

PaQ +

12 PQRS

gPQRS(x)aPa

RaSaQ

At each geometry, all matrix elements can be written as vacuumexpectation values of strings of operators. According to Wicks

theorem, only totally contracted terms contribute, depending

only on the overlap between the orbitals. Since the OMOs areorthormal at all geometries, the vacuum expectation values are

independent of the geometry.

The geometry dependence of the Hamiltonian is isolated in the

integrals.


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First derivative of the one-electron Hamiltonian

Consider

x PQ hPQ(x)aPaQ =

x PQ S1/2(x)h(x)S1/2(x)PQ aPaQ S(x) and h(x) are the overlap and Hamiltonian matrices at

the new geometry x0 +x in the basis of the UMOs.

When we expand around x0, we get

h(x0 +x) = h(0)(x0) + h

(1)(x0)x + . . .

S(x0 +x) = 1+ S(1)(x0)x + . . .

S

1/2(x0 +x) = 1 12S(1)(x0)x + . . .

where h(1)(x0) and S(1)(x0) are the first derivatives of h

and S in the UMO basis, computed at the reference

geometry x0.

One-index transformations Hence,

hPQ(x)

x

x=x0

=h(1)(x0)

12S

(1)(x0)h(0)(x0)

12h

(0)(x0)S(1)(x0)

We may write this in a compact brace notation for one-index

transformations,

hPQ(x)

x

x=x0

= h(1)PQ = h

(1)PQ

12

S(1), h(0)

PQ

where

{A, B}PQ =T

(APTBTQ + A

QTBPT)

{A, B}PQRS = T

(APTBTQRS + AQTBPTRS

+ ARTBPQTS + A

STBPQRT)


20/31

The HartreeFock gradient

With the AO-dependence isolated in the integrals of thesecond-quantization Hamiltonian, we may write the

HartreeFock gradient as

E(1)HF = E

(1)nuc +

PQ

h(1)PQD

HFPQ +

12

PQRS

g(1)PQRS d

HFPQRS

= E(1)nuc +I

h(1)II +

12

IJ

g(1)IIJJ g

(1)IJJI

= E(1)nuc +

I

h(1)II

IT

h(0)TIS

(1)TI

+ 12 IJ g(1)IIJJ g

(1)IJJI IJPg

(0)TIJJ g

(0)TJJIS

(1)TI

= E(1)nuc +I

h(1)II +

12

IJ

g(1)IIJJ g

(1)IJJI

IT

f(0)TIS

(1)TI

= E(1)nuc +I

h(1)II

(0)I S

(1)II

+ 12

IJ

g(1)IIJJ g

(1)IJJI

Parametrization of the HartreeFock state

The HF orbitals are obtained from the OMOs by a unitary

(or orhogonal) transformation,

CHF(x) = COMO(x)U(x), with U(x0) = 1

We can writeU

(x) = exp(), with

= . In second quantization, this translates into

=PQ

PQ aPaQ,

=

Spin and spatial restrictions may apply. In closed-shell HF

theory, one usually writes

= p>q

pq

(apaq aqap) =

p>q

pqEpq


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Orbitals at the displaced geometry

Geometry Orbital

x0 aP|vac =

HF

P

x0 +x aP|vac = OMO

P

x0 +x exp()OMO

P = exp()aP|vac =

HF

P

At the new geometry x0 +x, the HF orbital HF

P orbital is

replaced by HFQ through

HFQ = exp()aQaPa

P|vac

= exp()aQaP exp()exp()aP|vac

Thus, the replacement operator is

exp()aQaP exp()

CC energy at the displaced geometry

The CC energy at the displaced geometry is written as

ECC = OMO| exp(T)exp()Hexp() exp(T)|OMO

The wavefunction parameters in and T dodepend on the geometry.

The change of the AO basis is accounted for in H.

In the following, we shall write

ECC = 0| exp(T)exp()Hexp() exp(T)|0

with the Fermi vacuum |0 |HF OMO at the

reference geometry and expansion point x0.


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The closed-shell CC Lagrangian

We are now in the position to write the CC Lagrangian as

LCC = 0| exp(T)exp()Hexp() exp(T)|0

+

t| exp(T)exp()Hexp()exp(T)|0

+pq

pq(Fpq pqp)

where we have introduced the canonical condition, which

helps to implement the frozen-core approximation.

The orbital energies p are treated as wavefuntion

parameters. Derivatives of p are not required according to

the 2n+1 rule.

The closed-shell CC gradient

The CC gradient E(1)CC can be written as

E(1)CC = E

(1)nuc +

pq

h(1)pq Deff

pq +12

pqrs

g(1)pqrs deff

pqrs

= E(1)nuc +pq

h(1)pq Deffpq + 12 pqrs

g(1)pqrs d effpqrs

pq

S(1)pq Feff

pq

where we have introduced effective densitiesDeffpq and deffpqrs

and the effective Fock matrix

Feff

pq = o

Deff

pohoq +ors

deff

porsgqors


23/31

Effective CC densities

The effective CC densities contain the CC Lagrangian

densities plus contributions from the orbital rotation

multipliers pq,

Deffpq = |Epq|CC + pq

d effpqrs = |epqrs|CC + 2pqD

HF

rs prD

HF

qs

with pq =12(1 + pq)pq.

The effective CC densities depend on the zeroth-order

wavefunction parameters and multipliers. The zeroth-order wavefunction parameters and multipliers

are obtained by making the Lagrangian stationary.

Coupled-perturbed HartreeFock (CPHF)

The diagonal zeroth-order orbital rotation multipliers are

obtained from requiring that L/p = 0.

The off-diagonal zeroth-order orbital rotation multipliers are

obtained from the CPHF or Z-vector equations, whichfollow from L/rs = 0 for all r > s,

pq

pqApqrs + |[H, Ers]|CC = 0

with

Apqrs =

0|[ap, [aq, [E

rs, H]]]+|0


24/31

Second derivatives

Wavefunction parameters follow the 2n+1 rule.

Multipliers follow the 2n+2 rule (since the Lagrangian islinear in the multipliers).

Hence, we need the first-order wavefunction parameters

(amplitudes and orbital rotation parameters) to compute

second derivatives, but only zeroth-order multipliers,

(2) = E(20) + 2E(11)(1) + E(02){(1)}2

+(0) e(20) + 2e(11)(1) + e(02){(1)}2

E(mn) =m+nE

xmn, e(mn) =

m+ne

xmn

The 2n+1 and 2n+2 rules

The Lagrangian is written as

L = E+ e

where E is the energy and e the constraint e = 0.

For the first derivative, we obtain

dL

dx=

L

x= E(10) + (0)e(10)

The second derivative is obtained from

d

dx E(10) + E(01)(1) + (0)e(10) + (0)e(01)(1) + (1)e(00)

Note that e(00) = 0 and E(01) + (0)e(01) = L/ = 0.


25/31

The 2n+1 and 2n+2 rules

The first-order response of the wavefunction parameters is

obtained from requiring that de/dx = 0. This yields

e(10) + e(01)(1) = 0

The second derivative yields

(2) = E(20) + E(11)(1) + (0)e(20) + (0)e(11)(1) + (1)e(10)

+ (2)e(00) + E(01)(2) + (0)e(10)(2)

+ E(11)(1) + E(02)(1)

2+ (0)e(11)(1)

+

(0)

e(02) (1)2 + (1)e(01)(1)

= E(20) + 2E(11)(1) + E(02)(1)

2+ (0)

e(20) + 2e(11)(1) + e(02)

(1)

2

The symmetric approach

Thus far, we have used the symmetric formulafor second

derivatives w.r.t. 2 perturbations x and y

2E(11)(1) 2E

x

y+

2E

y

x In order to compute the second derivatives (e.g., the

molecular Hessian) of the CC energy, we need to solve

E(01) + (0)e(01) = 0

e(10) + e(01)(1) = 0

The zeroth-order multipliers equation is independent of the

perturbation, whereas the first-order wavefunction

parameters are determined by a set of equations that

involve the perturbation-dependent e(10) (w.r.t. x and y).


26/31

Dalgarnos interchange theorem

The asymmetric formula is obtained by considering thetotal derivative of the gradient,

(2) =d

dx

Ly

=

d

dx

E(010) + (00)e(010)

= E(110) + E(011)(10) + (10)e(010)

+ (00)e(110) + (00)e(011)(10)

with

E(klm) =k+l+mE

xkylm, (mn) =

m+n

xmyn, etc.

For mixed second derivatives (NMR chemical shifts, IR

intensities) it is sufficient to consider only the responses

(10) = /x and (10) = /x.

Time-dependent perturbations

In the following, we shall investigate a time-dependent

Hamiltonian of the form

H(t, ) = H(0) + V(t, )

where H(0) is the unperturbed molecular Hamiltonian andV(t, ) the time-dependent perturbation, written as sum ofFourier components

V(t, ) =N

j=N

Xjj(j)exp(ijt)

The Xj

are time-independent Hermitian operators,

j = j , and j(j) = j(j). Thus, V(t, ) is

Hermitian.


27/31

Frequency-dependent response functions

The time evolution of the observable A can be expressedby means of response functions,

A(t) = A0 +j

A; Xjje

i

jtj(j)

+ 12

jk

A; Xj , Xkj ,kei(j+k)tj(j)k(k) + . . .

Examples include the (frequency-dependent) polarizability

x; y and the first hyperpolarizability x; y, z1,2 .

Important symmetries:

A; B, C , . . .B,C ,... = A; C, B , . . .C ,B,...

= B; A, C , . . .(B+C+... ),C ,...

A; B, C , . . .B,C ,... = A; B, C , . . .

B,C ,...

Time-dependent Schrdinger equation

We write the time-dependent wavefunction |0 in thephase-isolated form

|0 = eiF(t)|0

Note that also |0 is time-dependent.

The time-dependent Schrdinger equation becomesH(t) i/t F(t)/t

|0 = 0

Projection onto 0| yields

F(t)t

Q(t) = 0|H(t) i/t |0


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Time-dependent quasi-energy

We term Q(t) the time-dependent quasi-energy. Note thatin the time-independent limit, F(t) = Et and Q(t) = E.

In CC response theory, we write |0 as

|0 |CC(t, ) = exp{T(t, )}|HF All time-dependence is contained in the cluster operator

T(t, ) (cf. orbital-unrelaxed properties).

The CC Lagrangian is

L(t, ) = (t, ) H(t) i/t CC(t, )(t, )| = HF| +

t(t, )| exp{T(t, )}

The FrenkelDirac variational principle

In the spirit of the FrenkelDirac variational principle

|H(t) i/t| = 0

we project the time-dependent Schrdinger equation

onto HF| and the excitation manifold | exp{T(t, )},

Q(t) = HF|H(t)|CC(t, )0 = | exp{T(t, )}

H(t) i/t

|CC(t, )

The CC equations can be written as

(t, ) it(t, )

t= 0

(t, ) = | exp{T(t, )}H(t)|CC(t, )


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Response functions

The response functions are defined as the nth derivative ofthe CC Lagrangian,

X1; X2, . . . , Xn2,...,n =12 C

dnL(t, )d1(1)d2(2) . . . dn(n)

with

Cf(1,2, . . . ,n) = f(1,2, . . . ,n)+f(1, 2, . . . , n)

The cluster amplitudes are expanded in the Fouriercomponents of the perturbation(s),

t(t, ) = t(0) +

j

tXj (j)j(j)eijt

+ 12

jk

tXjXk (j ,k)j(j)k(k)ei(j+k)t + . . .

The CC Jacobian

The amplitude responses are obtained from

( 1) tX1...Xn(1, . . . ,n) = X1...Xn(1, . . . ,n)

with = 1 + + n and

X1...Xn (1, . . . ,n) =n+1L(t, )

t1(1)2(2) . . . n(n)

is the Jacobian of the unperturbed system.

Similar equations determine the multiplier responses,

tX1...Xn(1, . . . ,n) (+ 1) =

X1...Xn(1, . . . ,n)


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Linear-response equations

Consider the linear response function X; Y.

The amplitude and multiplier responses are obtained from

the equations

( 1) tY() = Y()

tY() (+ 1) =

Y() =

Y() + FtY()

with

Y () =2L(t, )

tY(), Y () =

2L(t, )

tY(), F() =

2L(t, )

tt

The same equations are obtained for orbital-unrelaxed

second derivatives except for the 1 level shifts, whichare due to the term it(t, )/t in the amplitudesequation.

Linear-response functions

Recall the (symmetric) expression for the static 2nd

derivative,

(2) = E(20) + 2E(11)(1) + E(02){(1)}2

+ (0) e(20) + 2e(11)(1) + e(02){(1)}2= L(20) + 2L(11)(1) + L(02){(1)}2

The frequency-dependent polarizability becomes

xy(,) = x; y =12 C

{tx()y()

+ ty()x() + tx()Fty()}

= 12 Cty()x() + tx()

y()

The asymmetric formula is

x; y =12 C

{ty()x() + ty()x()}


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Poles and residues

( 1) tY() = Y()

tY

() (

+ 1

) =

Y

() = Y() + FtY() The response equations become singular when is

equal to an eigenvalue of the CC Jacobian. Thus, these

eigenvalues refer to an excitation energy.

The residues are related to transition moments.

The 2n+1 and 2n+2 rules apply.

x; y0 is the orbital-unrelaxed static polarizability. Frequency-dependent properties are obtained by

level-shifting the Jacobian in the response equations that

determine the perturbed amplitudes and multipliers.

Date post:	06-Apr-2018
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