Regarding Spin orbit coupling

[ajay kumar]

Respected Sir

I am conducting spin-orbit coupling calculations, focusing on four doublet states. To analyze the results, I have employed the HLSMAT and TRANLS keywords in my calculations.

1. HLSMAT Output: The results include eight eigenstates. Could you clarify the physical meaning of these states in the context of my calculation?
2. TRANLS Output: The output does not display diagonal or off-diagonal terms as expected. Could you identify any potential errors or omissions in my methodology?

I will attach both the input and output files for your reference. Your guidance in addressing these issues would be highly appreciated.

  

Thanking You

Yarram Ajay Kumar

[Peter Knowles]

Regarding question 1: The eigenstates are the eigenvectors of the (spin-orbit part of the) Breit-Pauli hamiltonian in the basis of your four Schrödinger states * [ms=1/2, ms=-1/2]. For your system, it looks as if the spin-orbit coupling effect is rather small.

It's difficult to comment on question 2, as it's not obvious what you are trying to achieve. 

Peter

On 23 Jan 2025, at 06:44, ajay kumar <ajay...@gmail.com> wrote:

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<ls-4-ci-dz-106.txt><ls-4-ci-dz-106.out>

[ajay kumar]

Thank you for your reply Sir. In the first part, there are eight eigen values, each value corresponds to [ms=+1/2 and ms=-1/2] and why are these values same for +1/2 and -1/2? The eigen value that's mentioned in the output is the total energy of each state [E_(ci)+E_(so)] and would i have to subtract it with mrci energies, to obtain Spin-orbit energies.

I would like to reiterate the second part of the question which is related to TRANLS. I would like to calculate the off-diagonal elements in the spin-orbit Hamiltonian.  Hence TRANLS have been used.

Thanking in advance and expecting a quick reply.

Yours Sincerely

    Ajay

[Peter Knowles]

No, it's not true that you can assign M_S values to those states, as the Breit Pauli operator does not commute with S_z, even though they still appear as degenerate Kramers pairs. You can see this in the eigenvectors with print,vls=0. In the output, the spin quantum numbers label the basis states, not the eigenvectors.

The program does already print the spin-orbit energies with respect to three origins, namely absolute energy, energy relative to lowest Schrödinger MRCI state, and lowest final eigenvalue.

Peter

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[ajay kumar]

Sir, i am  quite confused about what you have replied back. In the output file , they have mentioned about Spin-orbit eigen states and there are 8 eigen states and they are degenerate pairs,  what are these pairs and are these pairs diagonal elements? There Is E, E0 and E(1) what do these three entail? 

Secondly, how can we get off-diagonal elements of the spin orbit Hamiltonian.  

Thanking in advance. Expecting a quick reply.

Yours Sincerely

      Ajay

[Peter Knowles]

 As I said, and as you can read at https://www.molpro.net/manual/doku.php?id=spin-orbit-coupling, you can see the eigenvectors with print,vls=0. You need to read a textbook to understand the degeneracy in terms of Kramers' theorem.

From your example:

!MRCI STATE  1.4 Energy            -1155.973152229158

Spin-orbit eigenstates   (energies)

 ======================

   Nr         E             E-E0         E-E0           E-E(1)      E-E(1)      E-E(1)

            (au)            (au)        (cm-1)           (au)       (cm-1)        (eV)

   1 -1155.97315543    -0.00000320       -0.70      0.00000000        0.00      0.0000

   2 -1155.97315543    -0.00000320       -0.70      0.00000000        0.00      0.0000

   3 -1155.95093876     0.02221347     4875.29      0.02221666     4875.99      0.6045

   4 -1155.95093876     0.02221347     4875.29      0.02221666     4875.99      0.6045

   5 -1155.92216399     0.05098824    11190.62      0.05099143    11191.33      1.3875

   6 -1155.92216399     0.05098824    11190.62      0.05099143    11191.33      1.3875

   7 -1155.73392421     0.23922802    52504.48      0.23923121    52505.18      6.5098

   8 -1155.73392421     0.23922802    52504.48      0.23923121    52505.18      6.5098

So pretty obviously E0 is the lowest non-relativistic MRCI energy, and E(1) is the lowest spin-orbit energy.

To get the hamiltonian matrix, https://www.molpro.net/manual/doku.php?id=spin-orbit-coupling says print,hls=0.

Peter

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[ajay kumar]

Thank you sir. I will get back to you after doing further calculations and reading.

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