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#1 2024-08-21 06:40:04

chrotteg
Member
Registered: 2023-05-25
Posts: 4

Difference between i=j transition dipole moment and dipole moment?

Hi Professor Lu,

I was curious as to what the difference is regarding calculating the transition dipole moment for S1 to S1 versus the overall dipole moment of S1? I have gathered from other posts that there is a difference between the transition dipole moment and the excited state dipole moment. The issue I have run across is that the multiwfn output matches my gaussian output for the transition dipoles (confirmed by function 8, subfuntion 5, then 2, outputting transdipmom.txt), but I can't get multiwfn to match the dipole moment I get in gaussian when I do density=current. To do this I was assuming that in transdipmom.txt when i=j, S1 to S1 for example, it would be equivalent to the dipole moment. I also believed it may be the nuclear charge causing the issue so I checked dipmom.txt, however, my nuclear charge seemed negligable so S1 was pretty much the same value. I did lots to figure out the difference myself but wasn't able to, so I hope its not something small. Here is some of the values I am looking at.

PS: I did read that diffuse functions should not be used, I tried without them but still have the same issue.

Gaussian input:
# td=(nstates=15,root=1) rb3lyp/6-311++g(3d2f,3p2d) pop=full
density=current geom=connectivity 5d gfprint iop(9/40=5)


Guassian dipole moment output:
Dipole moment (field-independent basis, Debye):
    X=              0.1052    Y=              0.0562    Z=              0.0000  Tot=              0.1193
I know the values are small but I just worked on S1 for a simple example to figure out.


Gaussian Transition Dipole output:
Ground to excited state transition electric dipole moments (Au):
       state          X           Y           Z        Dip. S.      Osc.
         1         0.0000      0.0000     -0.0171      0.0003      0.0000
         2         0.0682     -0.0366      0.0000      0.0060      0.0008
         3        -0.0000     -0.0000      0.4373      0.1912      0.0309


Multiwfn transition dipole output from transdipmom.txt:
Ground state dipole moment in X,Y,Z:    1.000214    0.211523    0.000001 a.u.

Transition dipole moment between ground state (0) and excited states (a.u.)
     i     j         X             Y             Z        Diff.(eV)   Oscil.str
     0     1    -0.0000000     0.0000004    -0.0170737     5.59910     0.00004
     0     2     0.0682000    -0.0366168    -0.0000000     5.62580     0.00083
     0     3    -0.0000004    -0.0000002     0.4372912     6.59230     0.03088


Multiwfn i and j transition dipole moments:
Note: In below output the case of i=j corresponds to electronic contribution to dipole moment of excited state i
Transition dipole moment between excited states (a.u.):
     i     j         X             Y             Z        Diff.(eV)   Oscil.str
     1     1    -0.7586006    -0.2607146    -0.0000031     0.00000     0.00000


Multiwfn dipmom.txt output:
Excited state dipole moments (a.u.):
  State         X             Y             Z        exc.(eV)    exc.(nm)
     1     -0.758623     -0.260681      0.000001      5.5991      221.44

As can be seen, the S1 in dipmom.txt is no where near the gaussian dipole moment output, even if converted to Debye. The transition dipole moments are exact though. I am very curious as to what I am missing. Thank you so much for your time and help.

A little bonus question I thought of, what does the variation of the dipole moment respect to the ground state show thats different than the dipole moment? I understood the calculation of it but I was curious on the importance of it. Thank you again!

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#2 2024-08-21 16:23:29

sobereva
Tian Lu (Multiwfn developer)
From: Beijing
Registered: 2017-09-11
Posts: 1,830
Website

Re: Difference between i=j transition dipole moment and dipole moment?

The dipole moment obtained by Multiwfn in this way corresponds to unrelaxed excited density, while directly using "density" keyword will make Gaussian output dipole moment corresponding to relaxed excited density at final part of the output file. If you use "density=rhoci" instead, the dipole moment given by Gaussian will also correspond to unrelaxed excited density.

See example:

#p b3lyp/6-31G* iop(9/40=4) density=rhoci TD(root=2)

b3lyp/6-311G* opted

0 1
 C                  0.00000000    1.09275900    0.34582000
 C                  0.00000000    0.71582300   -0.95557500
 C                  0.00000000   -0.71582300   -0.95557500
 C                  0.00000000   -1.09275900    0.34582000
 O                  0.00000000    0.00000000    1.15706000
 H                  0.00000000    2.04717300    0.84155100
 H                  0.00000000    1.36964900   -1.81125900
 H                  0.00000000   -1.36964900   -1.81125900
 H                  0.00000000   -2.04717300    0.84155100

Finally Gaussian outputs

 Dipole moment (field-independent basis, Debye):
    X=              0.0000    Y=              0.0000    Z=             -1.0345  Tot=              1.0345

Multiwfn outputs

 Transition electric dipole moment between excited states (a.u.):
     i     j         X             Y             Z        Diff.(eV)   Oscil.str
     1     1     0.0000000    -0.0000000    -0.5354110     0.00000     0.00000
     1     2    -0.0000000    -1.6938922    -0.0000000     0.43340     0.03047
     1     3    -0.0308571    -0.0000000     0.0000000     1.57400     0.00004
     2     2     0.0000000    -0.0000000    -0.4084153     0.00000     0.00000
     2     3     0.0000000    -0.0000000    -0.0000000     1.14060     0.00000
     3     3     0.0000000     0.0000000     1.2196069     0.00000     0.00000

Pay attention to the Z component of the 2nd excited state given by Multiwfn, -0.4084153 a.u. *2.5417462 = -1.038 Debye, which is in good agreement with the -1.0345 printed by Gaussian.

However, if you use "density" rather than "density=rhoci", Gaussian prints

 Dipole moment (field-independent basis, Debye):
    X=              0.0000    Y=              0.0000    Z=             -0.8180  Tot=              0.8180

The result is not so close to Multiwfn.

By the way, when "nosymm" is not used, Gaussian automatically translates the system so that the origin is at center of nuclear charges, in this case contribution of nuclei to dipole moment is exactly zero, and thus you do not need to check the contribution.

Best regards,

Tian

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