Transition Dipole moment

[Ruairi Graham]

Dear Dirac experts,

I have done a KRCI calculation in linear symmetry and worked out tdms for:

.CIROOTS                                                                                            
1 3  
.CIROOTS
3 1 

The 1(2) state and the 3(1) state are both A2Pi states with the first one being A2Pi(1/2) and A2Pi(3/2). However, their tdm with the ground X2Sigma+ state are very different. 

The tdm section from one of the output files is:

 <initial state   |x,y,z|   final state>              TDM [a.u.] (x, y, z)           Norm [a.u.]  Norm [Debye]

  __________________       _________________    _______________________________________________________________

   * state  1(1   )         * state  2(1   )     0.00000000  0.00000000  0.11588854    0.11588854    0.29455928

   * state  1(1   )         * state  3(1   )     0.00000000  0.00000000  1.76245984    1.76245984    4.47972609

   * state  1(1   )         * state  1(3   )     1.75243362 -1.75243362  0.00000000    2.47831540    6.29924942

The tdm is small for the 2(1) state across all geometries not just this one. I assuming since the A²Π1⁄₂ state is omega = 1/2 this explains why there is only a small dipole moment with the ground state along the bonding axis (z axis) but for a Σ to Π transition should the dipole moments instead be along the perpendicular (x and y) axes? There is no way to specify in the input that this 2(1) state is a Π state as you can only specify the omega value. Is there a way around this as the tdm for the A²Π₃⁄₂ seems to be quite accurate as I have worked out the lifetime and it was very close to the literature value for this molecule. The literature value for the A²Π1⁄₂ is very similar to the 3/2 state but I have calculated a much longer lifetime due to the low tdm values in my output file. 

[peterso...@gmail.com]

I'm sure you've already checked this, but are  you sure the Pi_1/2 state is not the 3rd root in your KRCI and not the 2nd root, i.e., there is a low-lying Omega=1/2 between your ground state and the exited Pi?

regards, -Kirk

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[Ruairi Graham]

Hi Kirk,

Thank you for such a quick response. The following is in my output file:

    state #        symmetry      <omega>      <s_z>     <l_z>
  ----------     -----------   ------------  -------  --------

          1        1               1/2        0.5000    0.0000
          2        1               1/2       -0.4999    0.9999
          3        1               1/2        0.5000    0.0000
          1        3               3/2        0.5000    1.0000

So the 2(1) state is a Pi state. This is a complete guess but maybe DIRAC only computes tdms along the z axis for omega = 1/2 states? 

Regards<
Ruairi 

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[peterso...@gmail.com]

Dear Ruairi,

ok, I had a feeling that would be too easy :^)    I really doubt that Dirac has any restrictions on what components it computes.

So a Sigma-Pi transition is perpendicular (Delta_Lamba=±1) but an Omega=1/2 - 1/2 is parallel. So I can see that the tdm might be small and it gets its intensity via spin-orbit coupling. The transition to the Sigma state is a fully allowed parallel transition, so it makes sense its tdm is relatively large (with components along x and  y). What am I missing here?

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[peterso...@gmail.com]

Hi Ruairi,

as usual I replied much too quickly to this. After a chat with an actual card-carrying spectroscopist (name withheld since I might still mess things up), I found that I neglected a possibility that does access the 2Pi_1/2 state without resorting to SO coupling:

your perpendicular transition accessing the 2Pi_3/2:

2SIGMA(M_L=0, M_S = +1/2, M_J=+1/2) to 2Pi (M_L=+1, M_S=+1/2, M_J = +3/2)  Delta_Omega=+1

the parallel transition I was thinking of that accesses the 2Pi_1/2 via SO:

2SIGMA(M_L=0, M_S = +1/2, M_J=+1/2) to 2Pi (M_L=+1, M_S=-1/2, M_J = +1/2)  Delta_Omega=0

but there is this perpendicular transition that accesses the 2Pi_1/2 and should have the same TDM as the one accessing the 2Pi_3/2:

2SIGMA(M_L=0, M_S = +1/2, M_J=+1/2) to 2Pi (M_L=-1, M_S=+1/2, M_J = -1/2)   Delta_Omega=-1

Maybe Dirac ignores this last one since it is the same as the first and then only shows the parallel transition?

regards,  -Kirk

From: dirac...@googlegroups.com <dirac...@googlegroups.com> on behalf of Ruairi Graham <ruairig...@gmail.com>

Date: Thursday, February 26, 2026 at 2:02 PM
To: dirac...@googlegroups.com <dirac...@googlegroups.com>
Subject: Re: [dirac-users] Transition Dipole moment

To view this discussion visit https://groups.google.com/d/msgid/dirac-users/CAFO5yz4gpzJ8EPsMZ3M%2BhgRw8Su3ZRTY%3DEHvKN-F%2BsH6e4wU_g%40mail.gmail.com.