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Simplified Equation of MotionCoupled Cluster for Excited States

Joshua GoingsDepartment of Chemistry

University of Washington, Seattle jjgoings@uw.edu jjgoings

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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.

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Helgaker, Trygve, et al. “Recent advances in wave function-based methodsof molecular-property calculations.” Chemical Reviews 112.1 (2012): 543-631.

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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⟩

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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.

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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

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Shavitt, Isaiah, and Rodney J. Bartlett. Many-body methods in chemistryand physics: MBPT and coupled-cluster theory. Cambridge UniversityPress, 2009.

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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.

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Crawford, T. Daniel, and H. F. Schaefer. “An introduction to coupled clus-ter theory for computational chemists.” Reviews in computational chemistry14 (2000): 33-136.

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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⟩

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Bartlett, Rodney J. “Coupled cluster theory and its equation-of-motionextensions.” Wiley Interdisciplinary Reviews: Computational Molecular Sci-ence 2.1 (2012): 126-138.

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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.

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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) + · · ·

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(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)

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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!)

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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!)

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Why is this method promising?▶ Non-iterative solution for amplitudes▶ Far smaller prefactor▶ Lowers scaling by a factor

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Why might this method fail?▶ Total neglect of single excitations (Thouless 1960)▶ Neglect of higher excitation character▶ Cost/benefit of accuracy versus speed

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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

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