VMC and MBGFT

[martinspenke]

Dear Daniele,

1) I have seen in your paper : " Ab initio geometry and bright excitation of carotenoids: Quantum Monte Carlo and Many Body Green’s Function Theory calculations on peridinin"
that you assign symmetries to the excited states, namely Bu plus and Ag minus like symmetries.
I would like to know how you get these symmetries for the excited states because neither QE nor Abinit gives the symmetries of the orbitals participating in those specific transitions !?
I know that the excited state symmetry is a multiplication of symmetries of in that transition participating orbitals, but i have no idea how to extract the symmetries of the participating orbitals it self since QE and Abinit give both no information about participating orbital symmetries ? So how is it been done here in this paper ?

2) Assuming one has the excited state geometry of a single isolated molecule the calculation of its emission spectrum is the same as absorption spectrum calculation ?
(because i do not see any keyword specifying emission spectra calculations.)
If yes, it means i have to put this excited state geometry to abinit as it were the ground state to get KSS file and calculate in yambo absorption spectrum which in reality is a emission spectrum ? Is it correct how i am thinking ?
If no, how can i generally calculate emission spectra within yambo ?

3) As much as i looked in the tutorials and forum it seems that BSE calculations are unfortunately totally static,
so will there be in the next release a fully frequency dependent BSE as it is described in this paper : Double excitations in correlated systems: A many–body approach ?
Best Regards
Martin

[Daniele Varsano]

Dear Martin,

1) The symmetry of the excitations can be deduced by the symmetry of the orbitals participating in the excitation. ypp -a tells you the KS states participating. Of course if the molecule has some symmetries. In the paper you mention, it is caretonoid but not fully symmetric, this is why we call them Ag-like, Bu-like.


2) Correct, of course the difficult part is to know the excited state relaxed geometry

3) Actually this part of the code it is not included in the yambo release. It is our intension to resume it and include in the trunk source. Honestly I cannot say now when it will happen, but I hope soon.

Best,

Daniele

[martinspenke]

Dear Daniele,

1) Yes, i know but i do not know how to get the symmetries of each of these orbitals which are participating in the transition.
I have in the output file some thing like this :

Code: Select all

#  Band_V     Band_C     K  ibz     Symm.      Weight     Energy
  23.00000   39.00000    1.00000    1.00000    0.11681    7.52609
  23.00000   40.00000    2.00000    1.00000    0.09733    7.58538
  23.00000   40.00000    2.00000    5.00000    0.09733    7.58538
  23.00000   25.00000    1.00000    1.00000    0.07902    4.15276
  23.00000   25.00000    2.00000    1.00000    0.07837    4.15278

but i have no clue what the 4th column (Symm.) is ? What do these numbers in the 4th column (Symm.) really mean ? (1 1 5 1 1) ?
(I know these numbers are related to the symmetries but which symmetries exactly ?)

2) I was here thinking of CHAMP code by Prof. Filippi which should be able to give highly accurate excited state geometries. And then doing what i wrote before.

3) Hopefully BSE with dynamic capability will be released soon.

Regards
Martin

[Daniele Varsano]

Dear Martin,
the Symm. column tells you the index of the symmetry of your system. The symmetry transformation matrix of the system and relative indexes are listed in the report files:

Code: Select all

 [02.02] Symmetries

Please note that this is not related with the symmetry of the excitations, this gives you the K point in the entire BZ expanded by symmetry operation from the Kpoint of the irreducible Brilluoin zone (K ibz).

Best,
Daniele