Author Topic: fermi and triplet occupation  (Read 6975 times)

moere

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fermi and triplet occupation
« on: June 22, 2010, 09:43:53 PM »
Hi there,

I am trying to optimize a structure of a complex (with Ni)
in its triplet state. (using dft/pbe,svp)

So what I did was a eht with establishing a triplet occupation.

The geometry optimization resulted in a negative HOMO-LUMO Gap, according to "eiger".

Then I added
$fermi tmstrt=350.00 tmend=150.00 tmfac=0.980 hlcrt=1.0E-01 stop=1.0E-03 nue=2
to the control file (I found in in the forum and it already helped me for other problems)
and run a singlepoint.
Running eiger again resulted in having no negative HOMO-LUMO Gap anymore.

But when I looked up the orbitals in the control file it says I have 0.5 electrons in
one of the orbital.

Eiger says:
Code: [Select]
Total energy =  -2494.4305946070 H =    -67876.9486821 eV
 
 HOMO-LUMO Separation
   HOMO:   143. a  10 b1g   -0.15037398 H =     -4.09189 eV
   LUMO:   144. a   5 eg    -0.10644927 H =     -2.89663 eV
   Gap :                    +0.04392471 H =     +1.19525 eV
 
Number of MOs=    578, Electrons=    188.00, Symmetry: d4h 
 
   Nr.   Orbital    Occupation       Energy
  160. b  20 eu                    +0.050570 H =        +1.376 eV
  159. a  20 eu                    +0.050381 H =        +1.371 eV
  158. b  17 a1g                   +0.046001 H =        +1.252 eV
  157. a  17 a1g                   +0.045317 H =        +1.233 eV
  156. b   2 a1u                   +0.033028 H =        +0.899 eV
  155. a   2 a1u                   +0.032031 H =        +0.872 eV
  154. b  16 a1g                   +0.026476 H =        +0.720 eV
  153. a  16 a1g                   +0.024691 H =        +0.672 eV
  152. b   6 eg                    +0.022742 H =        +0.619 eV
  151. a   6 eg                    +0.020931 H =        +0.570 eV
  150. b   3 b2u                   +0.007446 H =        +0.203 eV
  149. a   3 b2u                   +0.005272 H =        +0.143 eV
  148. b   2 b1u                   -0.059352 H =        -1.615 eV
  147. a   2 b1u                   -0.060003 H =        -1.633 eV
  146. b  10 b1g                   -0.084731 H =        -2.306 eV
  145. b   5 eg                    -0.104036 H =        -2.831 eV
  144. a   5 eg                    -0.106449 H =        -2.897 eV
  143. a  10 b1g      1.000        -0.150374 H =        -4.092 eV
  142. b   4 eg       1.000        -0.172941 H =        -4.706 eV
  141. b   5 a2u      1.000        -0.181115 H =        -4.928 eV
  140. a   5 a2u      1.000        -0.182430 H =        -4.964 eV
  139. b   1 a1u      1.000        -0.187253 H =        -5.095 eV
  138. a   1 a1u      1.000        -0.187689 H =        -5.107 eV
  137. b  15 a1g      1.000        -0.193000 H =        -5.252 eV
  136. b   9 b2g      1.000        -0.201739 H =        -5.490 eV
  135. a   4 eg       2.000        -0.212609 H =        -5.785 eV
  134. b   2 b2u      1.000        -0.219447 H =        -5.971 eV
  133. a   2 b2u      1.000        -0.222073 H =        -6.043 eV
  132. a  15 a1g      1.000        -0.226051 H =        -6.151 eV
  131. b   3 eg       2.000        -0.231343 H =        -6.295 eV
  130. a   9 b2g      1.000        -0.231398 H =        -6.297 eV
  129. b   4 a2u      1.000        -0.232780 H =        -6.334 eV
  128. a   4 a2u      1.000        -0.234912 H =        -6.392 eV

and the control file says:

Code: [Select]
$alpha shells
 a1g     1-15                                   ( 1 )
 a2g     1-6                                    ( 1 )
 b1g     1-10                                   ( 1 )
 b2g     1-9                                    ( 1 )
 eg      1-4                                    ( 1 )
 a1u     1                                      ( 1 )
 a2u     1-5                                    ( 1 )
 b1u     1                                      ( 1 )
 b2u     1-2                                    ( 1 )
 eu      1-19                                   ( 1 )
$beta shells
 a1g     1-15                                   ( 1 )
 a2g     1-6                                    ( 1 )
 b1g     1-9                                    ( 1 )
 b2g     1-9                                    ( 1 )
 eg      1-3                                    ( 1 )
 eg      4                                      ( 0.5 )
 a1u     1                                      ( 1 )
 a2u     1-5                                    ( 1 )
 b1u     1                                      ( 1 )
 b2u     1-2                                    ( 1 )
 eu      1-19                                   ( 1 )

So I dont understand these lines in eiger:
135. a   4 eg       2.000
and
131. b   3 eg       2.000

And I dont understand the 0.5 in the control file.



And when I count all the electrons in the control file section I get 142.5
but eiger says there are 188 electrons. ???

I had something similar before but there I had two beta orbitals with 0.5 occupation
what at least made a bit sense as it still results in 1 electron in total.

When I then delete the fermi again in the control file
and run a optimization job my structure converges.

But what can I do to get a proper triplet optimization without a 0.5 occupation?
I would expect to find a optimized structure with integer occupation numbers.
Or are those with non integer occupation numbers are proper results already?

The structure with the neg. homo-lumo gap and that one with the 0.5 occupation only differ by ~200 cal/mol. (The latter has a lower energy).

I read in the manual that I can adjust tmend to low temperature to obtain integer occupation numbers.
I set it to 10, but it didnt help, I still revieve 0.5 and 0.9999999756 and sth like that.

Thanks in advance,

moere

The structure I am trying to optimize in its triplett state:
Code: [Select]
37
Energy = -2494.430594607
C     2.8641677    1.1020613    0.0000000
N     2.0323199    0.0000000    0.0000000
C     4.2517123    0.6874939    0.0000000
C     4.2517123   -0.6874939    0.0000000
C     2.8641677   -1.1020613    0.0000000
C     2.4355508    2.4355508    0.0000000
C     1.1020613    2.8641677    0.0000000
C     0.6874939    4.2517123    0.0000000
C    -0.6874939    4.2517123    0.0000000
C    -1.1020613    2.8641677    0.0000000
N     0.0000000    2.0323199    0.0000000
Ni    0.0000000    0.0000000    0.0000000
N     0.0000000   -2.0323199    0.0000000
C    -1.1020613   -2.8641677    0.0000000
C    -0.6874939   -4.2517123    0.0000000
C     0.6874939   -4.2517123    0.0000000
C     1.1020613   -2.8641677    0.0000000
C    -2.4355508    2.4355508    0.0000000
C    -2.8641677    1.1020613    0.0000000
C    -4.2517123    0.6874939    0.0000000
C    -4.2517123   -0.6874939    0.0000000
C    -2.8641677   -1.1020613    0.0000000
N    -2.0323199    0.0000000    0.0000000
C    -2.4355508   -2.4355508    0.0000000
C     2.4355508   -2.4355508    0.0000000
H    -1.3698557    5.1126219    0.0000000
H     1.3698557    5.1126219    0.0000000
H    -5.1126219    1.3698557    0.0000000
H    -5.1126219   -1.3698557    0.0000000
H     1.3698557   -5.1126219    0.0000000
H    -1.3698557   -5.1126219    0.0000000
H     5.1126219    1.3698557    0.0000000
H     5.1126219   -1.3698557    0.0000000
H    -3.2143272   -3.2143272    0.0000000
H     3.2143272    3.2143272    0.0000000
H     3.2143272   -3.2143272    0.0000000
H    -3.2143272    3.2143272    0.0000000

Arnim

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Re: fermi and triplet occupation
« Reply #1 on: October 25, 2010, 01:10:46 PM »
Hi,

0.5 in an E-representation means that you have  2 and not 4 electrons in this shell. You can avoid this by using the next lower
Abelian group, D2h in this case.
Lowering the symmetry would also be a good idea, because otherwise no Jahn-Teller distortion would be possible.

Best regards,
Arnim