simplified equation of motion coupled cluster for excited ... › assets › eompt_plain.pdf ·...
TRANSCRIPT
..
Simplified Equation of MotionCoupled Cluster for Excited States
Joshua GoingsDepartment of Chemistry
University of Washington, Seattle [email protected] jjgoings
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
Molecular properties:Linear response (LR) or equation of motion (EOM)Calculation of (most) properties require:
▶ ..Response theory due to perturbation▶ ..Approximation to the wavefunction
The order in which theory is applied matters!..LR-CC = ..CC + ..LR..EOM-CC = ..LR + ..CC
We will focus on EOM methods.
..
Helgaker, Trygve, et al. “Recent advances in wave function-based methodsof molecular-property calculations.” Chemical Reviews 112.1 (2012): 543-631.
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
We form an exact excited state from the exact ground state
|Ψe⟩ = R|Ψg⟩
R = R1 + R2 + · · · = rai a†aai + rab
ij a†aa†bajai + · · ·
Likewise, we can generate the exact ground state
|Ψg⟩ = eT|Ψ0⟩
T = T1 + T2 + · · · = tai a†aai + tab
ij a†aa†bajai + · · ·
..HeTRm|Ψ0⟩ = EmeTRm|Ψ0⟩ or ..HRm|CC⟩ = EmRm|CC⟩
..
Stanton, John F., and Rodney J. Bartlett. “The equation of motion cou-pled cluster method. A systematic biorthogonal approach to molecularexcitation energies, transition probabilities, and excited state properties.”The Journal of chemical physics 98 (1993): 7029.
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
It is convienient to use the normal-ordered Hamiltonian, HN.
HN = H − ⟨0|H|0⟩
In other words, HN is now a “correlation operator”. In secondquantization this gives:
HN = fpq{a†paq}+1
4⟨pq||rs⟩{a†pa†qasar}
or simply..HN = F + V
..
Shavitt, Isaiah, and Rodney J. Bartlett. Many-body methods in chemistryand physics: MBPT and coupled-cluster theory. Cambridge UniversityPress, 2009.
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
In addition to normal ordered operators, we use Wick’s Theoremto simplify the CC equations.
H̄N = e−THNeT
= HN + [HN,T] +1
2[[HN,T],T] + · · · (infinitely more!)
· · · a bit of work · · ·
= HN + HNT1 + HNT2 +1
2HNT2
1 + · · · (only 11 more terms!)
= (HNeT)c
Turns out all the T operators must share an index with HN(“connected cluster”), and the expression truncates naturally.
..
Crawford, T. Daniel, and H. F. Schaefer. “An introduction to coupled clus-ter theory for computational chemists.” Reviews in computational chemistry14 (2000): 33-136.
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
One final result before we continue. We can solve for excitationsdirectly:
[H̄,Rm]|0⟩ = H̄Rm|0⟩ − RmH̄|0⟩= EmRm|0⟩ − E0Rm|0⟩= ωRm|0⟩
where ωm = Em − E0. Applying Wick’s theorem, we keepconnected R and H̄ terms, giving us a final expression of
..(H̄NR)c|0⟩ = ωR|0⟩
..
Bartlett, Rodney J. “Coupled cluster theory and its equation-of-motionextensions.” Wiley Interdisciplinary Reviews: Computational Molecular Sci-ence 2.1 (2012): 126-138.
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
Let’s use EOM-CCSD as an example.
T = T1 + T2
The actual solution requires diagonalizing H̄ in the space ofsingly and doubly excited determinants(
⟨Φai |H̄|Φc
k⟩ ⟨Φai |H̄|Φcd
kl ⟩⟨Φab
ij |H̄|Φck⟩ ⟨Φab
ij |H̄|Φcdkl ⟩
)The matrix elements are evaluated using diagrammatic tech-niques.
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
k c k c l d m e k ci
a
c
j b
ai
k
j bi a j
b
ci a b
j
ki
a
c
b
dj
ai
k
j
lb i a
k c
j b
ei
ka b
fj c
ima j
nb
ci
ma b
ej e
ik
a j
mb e
ima b
cj e
ima j
kb
le f jk a bi dm n bc i ja be m ak c
ji be
i
fk a j
m
be
a
nc i j
mne c jm a bi ke f jm a bi i
a
c
b
ej
ai
k
j
mb i
cma
j b
ie k
a
j b i a
e bk j
a i
m bc jk
a
ei
cj b
k
i
ma
cj b c
i
ma b
dj
ka
ei j
lb
ce
a
mi j b
ke
i
ma j b nc e jm a bi
fk m be i ja bc m an e
ji be k am f
ji bc
i
em a j
k
be
i
cn a j
mb
c
a
me i j
kb
e
a
kf i j
m da j bk c li
ab l dk c iji ac m j b
a ik e j b
m
a
ci
ej b
m
i
ka
ej b e
ik
a b
fj m
ac
i j
nb
ec
a
ki j b
mc
i
ka j b
bm d n
i j ac
bk f l
i j ae e
b i ak c m
j
a bk c
i d m j a bm e
i f k j i bk c
a l e j i bm e
a n c j
ia
ca
i
k
i a
k ci a
l d
ia
c l da
i
k l d
i a
ckmei a
fmke
i a
fnmc
ei
a
c m ea
i
k mi
a
e k c
ai
m k c
i ae k
i ac m
i a
lke d
a i
lcm d m e k ci a
i a
nmc e
a i
mek f
k c m ei a
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
The diagrams are numerous, and scale as bad as O(N6).We want to use the tools of perturbation theory to simplify theequations. Introducing a scalar ordering parameter λ
HN = F + λV
Similarly, we expand the T operator perturbatively
T = λT(1) + λ2T(2) + λ3T(3) + · · ·
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
(HNeT)c |Ψ0⟩ = HN |Φ0⟩+ HNT |Φ0⟩+1
2HNT2 |Φ0⟩+ · · ·
= (F + λV) |Φ0⟩+ (F + λV)(λT(1) + λ2T(2) + · · · ) |Φ0⟩
+1
2(F + λV)(λT(1) + λ2T(2) + · · · )2 |Φ0⟩+ · · ·
Collecting terms of like order λ yields, with H̄N = (HNeT)c
H̄(0)N = F
H̄(1)N = V + FT(1)
H̄(2)N = VT(1) + FT(2) +
1
2FT(1)T(1)
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
Unlike CCSD, we can solve for the T1 and T2 amplitudes di-rectly:
⟨Φai | H̄(1) |Φ0⟩ = 0
=∑
bfabtb(1)
i −∑
jfijta(1)
i
By the diagonal nature of the canonical Fock matrix elements,ta(1)i = 0. In a similar manner,
⟨Φabij |H̄(1) |Φ0⟩ = 0
= ⟨ij||ab⟩ − (fii + fjj − faa − fbb)tab(1)ij
tab(1)ij =
⟨ij||ab⟩ϵi + ϵj − ϵa − ϵb
For our reference, this gives the MP2 energy expression backdirectly (sanity check!)
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
Finally plugging in terms, we have three new methods to try
EOM-MBPT2 =
[⟨S| H̄(0→2) |S⟩ ⟨S| H̄(0→2) |D⟩⟨D| H̄(0→2) |S⟩ ⟨D| H̄(0→2) |D⟩
]
EOM-MBPT(D) =[
⟨S| H̄(0→2) |S⟩ ⟨S| H̄(0→2) |D⟩⟨D| H̄(0→2) |S⟩ ⟨D| ..H̄(0→1) |D⟩
]
EOM-MBPT(2) =[
⟨S| H̄(0→2) |S⟩ ⟨S| ..H̄(0→1) |D⟩⟨D| ..H̄(0→1) |S⟩ ⟨D| ..H̄(0→1) |D⟩
](In fact, we can derive CIS and the CIS(D) families of equationsthis way!)
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
Why is this method promising?▶ Non-iterative solution for amplitudes▶ Far smaller prefactor▶ Lowers scaling by a factor
..
Why might this method fail?▶ Total neglect of single excitations (Thouless 1960)▶ Neglect of higher excitation character▶ Cost/benefit of accuracy versus speed
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .
..
Why is this method promising?▶ Non-iterative solution for amplitudes▶ Far smaller prefactor▶ Lowers scaling by a factor
..
Why might this method fail?▶ Total neglect of single excitations (Thouless 1960)▶ Neglect of higher excitation character▶ Cost/benefit of accuracy versus speed
..... ..... ..... . ..................... ..................... ..................... ..... ..... . ..... ..... ..... .