EOM CCSD: how is Character of excitation defined?

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Nanna Falk Christensen

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Jan 24, 2022, 7:47:27 AMJan 24
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Hi Molpro users,

 I am running some EOM CCSD calculations, and I am trying to determine how much my calculated transitions are singly and doubly excited. 

Can anyone point me in a direction of a paper or some documentation, where the character of excitation is defined, and it is described how it is calculated?


I have so far not found anything in the molpro manual, or by googling this and related terms.


Thank you in advance, 

Nanna

Tatiana Korona

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Jan 25, 2022, 2:00:21 AMJan 25
to Nanna Falk Christensen, molpro-user
Dear Nanna,

This is all presented in the output. Look for:

   Final Results:
  ==============

  State    Exc. Energy    Total Energy     Weight(S)     Weight(D)      Delta E    Conv   Max. Coef.     Excitation

e.g.
1.2     0.27345717     -75.99513662     0.941E+00     0.590E-01     0.524E-08     Y     0.84803      1.2 ->  4.1

Means that this is the first state of second symmetry (1.2) of the excitation energy 0.27 a.u., with the weight of singles 0.941, and of doubles 0.059.
Then "Y" means the energy is converged (in some difficult cases the default number of iterations might be not sufficient), 0.848 is the maximal coefficient, and 1.2->4.1 - the corresponding CSF (the 1.2 spinorbital is replaced by 4.1 in the reference determinant).
Additionally, above the "Final Results" you will see a more detailed list of the largest coefficients (look for the string "Coefficient     Excitation").
 7029
Note that by default only excitation energies and rhs eigenvectors are calculated. In order to obtain e.g. transition moments, you should ask for this explicitly (see TRANS keyword in the manual). For properties you need both left- and right-hand side eigenvectors, so properties are twice as expensive as the energy only.

The general theory on EOM-CCSD is presented in Stanton and Bartlett paper (JCP, 98,  7029, 1993), while Molpro implementation in Korona and Werner, DOI: 10.1063/1.1537718 (although for the rhs part only).

Best wishes,

Tatiana





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Nanna Falk Christensen

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Jan 25, 2022, 9:18:19 AMJan 25
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Hi Tatiana, 
Thank you for your quick and thorough response. 
I have some additional questions:
1) Is it correctly understood that the weights are the sums of the amplitudes of singly excited and doubly excited slater determinants for a given excited state, assuming that they are normalized to one, or where do the weights come from?

2) As far as I have understood, calculation of singlet to triplet transition energies is not possible using EOM-CCSD in molpro. Is it just not implemented, or is there something inherent in the EOM formalism that inhibits the calculation of singlet to triplet transition energies, or have I simply missed the correct set of keywords for doing this?
I know that oscillator strengths are a different story, requiring the inclusion of spin orbit coupling, but would singlet to triplet transition energies be possible?
I have read your paper (JCP 2003), and Stanton and Bartlett 1993, and I could not read anywhere where it was stated that it was impossible, at least from my understanding.

Many thanks for your help,
Nanna 

Tatiana Korona

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Jan 25, 2022, 3:50:57 PMJan 25
to Nanna Falk Christensen, molpro-user
Dear Nanna,

1) We analyse single and double excitation amplitudes of the excitation operator, which created EOM rhs excited state from the CCSD reference: Psi = R exp(T) Phi0, i.e. we analyse "R". It is normalized before the analysis.
2) In Molpro singlet EOM-CCSD excitations from closed-shell reference are implemented. There are other implementations where excitations from ground-state singlet to triplet excited states are available, but not in Molpro.

Best wishes,

Tatiana


Nanna Falk Christensen

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Jan 26, 2022, 3:33:51 AMJan 26
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Thank you very much!
Nanna
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