[CP2K-user] [CP2K:21257] Broken Simmetry NEL

[Krack Matthias]

Dear Lorenzo

Sorry, my suggestion for the &BS section was wrong, try this instead

  &BS
    &ALPHA
      N    3  4
      L    2  0
      NEL  0 -1
    &END ALPHA
    &BETA
     N     3  4
     L     2  0
     NEL  -2 -1
    &END BETA
  &END BS

Best

Matthias

From: cp2k at googlegroups.com <cp2k at googlegroups.com> on behalf of lorenzo briccolani <lore.bricco.184 at gmail.com>
Date: Wednesday, 5 March 2025 at 19:54
To: cp2k at googlegroups.com <cp2k at googlegroups.com>
Subject: Re: [CP2K:21253] Broken Simmetry NEL
Dear Matthias,

Thank you very much for your help, as always—it is greatly appreciated.

Just to ensure I fully understand, the first column of values defines the target orbital, while the second column corresponds to the initial orbital. Therefore, in the case of the alpha electron, I add one electron with spin-up (NEL +1) in the 3d orbital by removing it from the 4s orbital. For the beta electron, I do not add any electron (NEL -1), transferring it from the 4s to the 3d orbital.

I apologize if I seem insistent, but I would like to clarify this point, as the manual refers to a single real value.

Following your recommendations, I assigned an initial multiplicity of two to the Cu atom. However, I am unsure why the calculation still assigns 10 valence electrons overall, while for the beta electrons, it assigns 4.5 instead of 4. Because the values enclosed in square brackets, I believe, refer to the number of alpha or beta electrons contained in each orbital. This is why, in the case of beta electrons, I expected D 4.00 instead of 4.50. Below, I have included the relevant lines from the output:
Guess for atomic kind: Cu

 Electronic structure
    Total number of core electrons                                         18.00
    Total number of valence electrons                                      10.00
    Total number of electrons                                              28.00
    Multiplicity                                                         doublet
    Alpha Electrons
    S   [  1.00  1.00  1.00]
    P   [  3.00  3.00]
    D      5.00
    Beta Electrons
    S   [  1.00  1.00  1.00]
    P   [  3.00  3.00]
    D      4.50

Best Regards

Lorenzo Briccolani

On Wed, 5 Mar 2025 at 12:02, Krack Matthias <matthias.krack at psi.ch<mailto:matthias.krack at psi.ch>> wrote:
Dear Lorenzo

The valence electron configuration of the Cu-q11 pseudopotential employed by you is 3d10 4s1. The following &BS section should create an initial 3d9 doublet configuration for a Cu(2+) atom:

 UKS
  MULTIPLICITY 2

  &BS
    &ALPHA
      N    3  4
      L    2  0
      NEL +1 -1
    &END ALPHA
    &BETA
     N    3  4
      L    2  0
      NEL -1 -1
    &END BETA
  &END BS
Note that these BS setting affects only the electron configuration of the initial atomic guess and the converged result might differ.

HTH

Matthias

From: cp2k at googlegroups.com<mailto:cp2k at googlegroups.com> <cp2k at googlegroups.com<mailto:cp2k at googlegroups.com>> on behalf of lorenzo briccolani <lore.bricco.184 at gmail.com<mailto:lore.bricco.184 at gmail.com>>
Date: Tuesday, 4 March 2025 at 19:29
To: cp2k <cp2k at googlegroups.com<mailto:cp2k at googlegroups.com>>
Subject: [CP2K:21245] Broken Simmetry NEL
Dear Users,

I am reaching out as I am uncertain about the correct assignment of NEL values in the BS routine. The system under investigation consists of a molecule with a central Cu2+ atom coordinating two ligands in a square planar geometry. In a standard spin density calculation (UPBE/DZVP-MOLOPT) with a multiplicity of two, the spin densities are well reproduced, with the unpaired electron localized precisely on the copper atom. The Hirshfeld output reports the following values:

Element  Kind  Ref Charge     Population       Spin moment  Net charge
           Cu     4      11.000        6.182   5.514            0.668     -0.695

My objective is to replicate this result using the BS approach. For a Cu2+ ion, I assume the configuration should be as follows:
&KIND Cu
     &BS ON
      &ALPHA
       L = 0 2
       N = 4 3
      NEL = -2 ?
      &END
      &BETA
       L = 0 2
       N = 4 3
      NEL = -2 ?
      &END
     &END

I have conducted several tests; however, I am unable to obtain the correct number of valence electrons and the expected multiplicity in the output. I have attached the input file for reference (opt.inp).

I would be sincerely grateful for any suggestions regarding the appropriate assignment of the NEL values.

Best regards,

Lorenzo Briccolani
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