Regarding the matter of "reordering"
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Mhmd Al ahmad
Apr 9, 2025, 7:55:51 AMApr 9
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Dear DIRAC experts
I wanted to obtain the orbital radius of the 6s¹ state in a gold atom in the absence of all electrons below the 6s orbital.
First case:
As a first step, I performed a calculation for a gold atom. I adjusted the occupancy to keep only one electron in the atom. Since the DIRAC program would place this electron in the 1s orbital, I performed a "reordering based on filling priority" to place the electron in the 6s orbital instead.
Second case:
As a first step, I performed a calculation for a hydrogen-like gold atom (a gold nucleus with one electron), then I performed a "reordering based on filling priority" to place the electron in the 6s orbital instead of 1s.
I initially thought these two procedures were equivalent, but the calculations showed otherwise—as if the "fingerprint" of the removed electrons in the first case had influenced the result.
My question is: Which problem is DIRAC solving in the first case? I mean, what is the difference analytically, at the level of Hamiltonian, between the two cases ?
Best regards,
Mohamed
Trond SAUE
Apr 9, 2025, 8:13:49 AMApr 9
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Dear Mohamed,
it is difficult to answer without seeing any output files.
Best regards,
Trond
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Trond Saue
Laboratoire de Chimie et Physique Quantiques
UMR 5626 CNRS --- Université Toulouse III-Paul Sabatier
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Book: Principles and Practices of Molecular Properties: Theory, Modeling, and Simulations
Trond Saue
Laboratoire de Chimie et Physique Quantiques
UMR 5626 CNRS --- Université Toulouse III-Paul Sabatier
118 route de Narbonne, F-31062 Toulouse, France
Phone : +33/561556361 Fax: +33/561556065
Mail : trond...@irsamc.ups-tlse.fr
Web : https://dirac.ups-tlse.fr/saue
DIRAC : http://www.diracprogram.org/
ESQC : http://www.esqc.org/
ERC Advanced Grant: HAMP-vQED
Book: Principles and Practices of Molecular Properties: Theory, Modeling, and Simulations
Mhmd Al ahmad
Apr 9, 2025, 10:20:42 AMApr 9
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I'm sorry for the oversight.
My input files are attached below.
Thank you for your effort and time,
Mohamed.Trond SAUE
Apr 9, 2025, 11:01:11 AMApr 9
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On 4/9/25 13:55, Mhmd Al ahmad wrote:
I wanted to obtain the orbital radius of the 6s¹ state in a gold atom in the absence of all electrons below the 6s orbital.
First case:
As a first step, I performed a calculation for a gold atom. I adjusted the occupancy to keep only one electron in the atom.
What do you mean by that ? In the output file Au_Au.out(1st case) you give the following occupation
*SCF .CLOSED SHE 40 38 .OPEN SHELL 1 1/2,0
so you end up doing a HF calculation with 78 electrons in closed shells and a single electron in an open shell. Note that DIRAC does not do open-shell HF, rather AOC HF,
see https://www.diracprogram.org/doc/release-25/tutorials/open_shell_scf/aoc.html#aoc
Looking a the Mulliken population analysis in the output we see that the HOMO is indeed 6s
* Electronic eigenvalue no. 21: -0.2177519800139 (Occupation : f = 0.5000) s 1/2; 1/2 ============================================================================================ * Gross populations greater than 0.01000 Gross Total | L Ag Au s -------------------------------------- alpha 1.0000 | 1.0000 beta 0.0000 | 0.0000
This calculation gives
Total energy : -19037.871907602974
ert
Since the DIRAC program would place this electron in the 1s orbital,
No, when doing AOC HF, orbitals are ordered by energy, starting with closed shells and ending with open shells, so you do get 6s half occupied.
Looking in the output file "Au_Au(1e- in the 6s orbital).out(1st case)" you have specified
**WAVE FUNCTION
.SCF
.REORDER MO'S
21,1..20
*SCF
#.CLOSED SHE
# 40 38
.OPEN SHELL
1
1/2,0
.OVLSEL
This means that you are calculating a one-electron gold atom. You are apparently starting from the orbitals obtained in the previous calculation since I find
* REACMO: Coefficients read from CHECKPOINT - Total energy: -19037.8719076028647
and do reorder them. From Mulliken population analysis we find
* Electronic eigenvalue no. 1: -24.654160810810 (Occupation : f = 0.5000) s 1/2; 1/2 ============================================================================================ * Gross populations greater than 0.01000 Gross Total | L Ag Au s -------------------------------------- alpha 0.9998 | 0.9997 beta 0.0002 | 0.0000
This calculation gives
Total energy : -3432.796909938835
Looking in the output file "Au_Au(1e- in the 6s orbital).out(2nd case)" you have specified
**WAVE FUNCTION
.SCF
.REORDER MO'S
21,1..20
*SCF
#.CLOSED SHE
# 40 38
.OPEN SHELL
1
1/2,0
.OVLSEL
so the same as before, but now I find
REACMO: Coefficients read from CHECKPOINT - Total energy: -3432.79690993896111
so now you are reordering the orbitals obtained in the in the output file "Au_Au(1e- in the 6s orbital).out(1st case)" and you get
* Electronic eigenvalue no. 1: -91.186084973548 (Occupation : f = 0.5000) s 1/2; 1/2 ============================================================================================ * Gross populations greater than 0.01000 Gross Total | L Ag Au s -------------------------------------- alpha 0.9984 | 0.9976 beta 0.0016 | 0.0000
as well as
Total energy : -91.531949642097928
Finally there is the output file "Au_Au(1e- in the 1s orbital).out(2nd case)". Here you specify
**WAVE FUNCTION
.SCF
#.REORDER MO'S
#21,1..20
#
*SCF
#.CLOSED SHE
# 40 38
.OPEN SHELL
1
1/2,0
so here you simply calculate a one-electron system starting from
scratch and get
* Electronic eigenvalue no. 1: -3418.6476725318 (Occupation : f = 0.5000) s 1/2; 1/2 ============================================================================================ * Gross populations greater than 0.01000 Gross Total | L Ag Au s B3uAu _small B2uAu _small B1uAu _small ----------------------------------------------------------------------------------- alpha 0.9391 | 0.9087 0.0000 0.0000 0.0304 beta 0.0609 | 0.0000 0.0304 0.0304 0.0000
I performed a "reordering based on filling priority" to place the electron in the 6s orbital instead.
Second case:
As a first step, I performed a calculation for a hydrogen-like gold atom (a gold nucleus with one electron), then I performed a "reordering based on filling priority" to place the electron in the 6s orbital instead of 1s.
I initially thought these two procedures were equivalent, but the calculations showed otherwise—as if the "fingerprint" of the removed electrons in the first case had influenced the result.
My question is: Which problem is DIRAC solving in the first case? I mean, what is the difference analytically, at the level of Hamiltonian, between the two cases ?
I fear that my overall conclusion is that you seem somewhat confused with respect to theory. What do you really want to calculate ?
A 6s-orbital of a one-electron gold atom or in a neutral gold atom ?
All the best,
Mhmd Al ahmad
Apr 9, 2025, 3:59:14 PMApr 9
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Dear Prof.,
Am trying to calculate the 6s orbital radius in the one-electron system Au^{78+}.
In the first procedure, I start from an input representing neutral Au atom (file: "Au_Au.out(1st case)"), to get as output "Au_Au(1e- in the 6s orbital).out(1st case)".
In the second procedure, I start from an input representing H-like Au ion, i.e. Au^{78+} (file: "Au_Au(1e- in the 1s orbital).out(2nd case)"), to get as output "Au_Au(1e- in the 6s orbital).out(2nd case)".
Why do I get different results in the two procedures?
Many thanks.
Am trying to calculate the 6s orbital radius in the one-electron system Au^{78+}.
In the first procedure, I start from an input representing neutral Au atom (file: "Au_Au.out(1st case)"), to get as output "Au_Au(1e- in the 6s orbital).out(1st case)".
In the second procedure, I start from an input representing H-like Au ion, i.e. Au^{78+} (file: "Au_Au(1e- in the 1s orbital).out(2nd case)"), to get as output "Au_Au(1e- in the 6s orbital).out(2nd case)".
Why do I get different results in the two procedures?
Many thanks.
Visscher, L. (Luuk)
Apr 10, 2025, 3:04:26 AMApr 10
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Dear Mohamed,
Why do you want to compute this with DIRAC ? You'll find the analytical expression for the mean radius as equation 7.48 in the book of Dyall and Faegri. See attach.
best regards,
Luuk
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Mhmd Al ahmad
Apr 10, 2025, 2:17:03 PMApr 10
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Dear Prof. Visscher,
What I would like to know, since DIRAC sometimes is like a blackbox where inner details are not clear, is what is the effect of "dropping" all the inner electrons -using the "reordering" instruction and filling just a higher orbital- on the Hamiltonian?
Normally, I would expect the same physics as an H-like atomic system where all expressions are known analytically. However, it seems to me the results are different, and my question is why ?
As an example, I took the radius of 6s orbital in the system Au^{78+}:
If I consider this as an H-like system, then I get a value of this radius.
However, if start with a neutral Au, then apply "reorder" and put just one electron on the 6s orbital, I find a value for the radius different from the H-like system. Why ?
Greetings,
Mohamed
Visscher, L. (Luuk)
Apr 11, 2025, 2:37:52 AMApr 11
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Dear Mohamed,
DIRAC uses a finite basis, so results should be different.
Please consult appropriate textbooks first before asking such elementary questions. This mailing list is meant for questions about the program, it is not a tsudy programme.
Luuk
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