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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.
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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
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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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Equation-of-motion CC theory (EOM-CC)
IN EOM-CC theory, we expand the excited states in the
space spanned by all |),
|ck) =
ck|) =
ck exp(T)|
=
ck exp(T)|HF =
ck exp(T)|HF
=
ck|CC = exp(T)
ck|HF
The EOM-CC excited state may be regarded as beinggenerated from a conventional expansion in Slater
determinants by the application of an exponential operator
containing the ground-state amplitudes.
The EOM-CC eigenvalue problem
In the biorthogonal basis, we may set up EOM-CC
wavefunctions of the form
|ck) = ck|), (ck| = c
k (|
and express the energy as a pseudo-expectation value
Ek = (ck|H|ck), with cTi cj = ij
For the ground state, we have c00 = 1 and c0
= 0 for > 0.Also, c00 = 1 and c
0 = t for > 0. Hence,
E0 = (c0|H|c0) = |H|CC = ECC,var
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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.
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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
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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)
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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
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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
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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.
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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).
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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.
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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.
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