Wondering about Unit conversion in RT-TDDFT


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I am interested in RT-TDDFT. I will use this method to calculate the UV-spectrum of a (huge) sensitizer molecule (about 100 atoms roughly). I guess that the calculation will heavily consume the computing time. That's why I need to know the possible maximum time step I can use. However, I am just curious about conversion of Unit for total time & time step and having two following questions.

Quantity Conversion
Time 1 au = 0.02419 fs
Length 1 au = 0.5292 A
Energy 1 au = 27.2114 eV
Electric field 1 au = 514.2 V/nm
Dipole moment 1 au = 2.542 D
Reference

The studied spectral bandwidth is 300 - 700 nm. Then I convert this range from nm to fs using a converter on this website click here., which gave me a resulting between 1.00 - 2.33 fs.

I selected 1.00 fs (300 nm), which has a highest energy in ev, as a ωmax

So, why is this time step value (in atomic unit) so large ?

And also if I want to calc spectrum to cover UV range, how much total time I should use ?

Thank you in advance for your kind comment and valuable answer.

Best,
Rangsiman

Forum Regular
You have a couple mistakes in your math, as well as a conceptual misstep.

First, the formula for maximum time step is pi over the maximum angular frequency. You have converted from wavelength to period and then have used the period in the formula. The angular frequency corresponding to a period of 1 fs is 2*pi fs^{-1}. So the maximum time step you should have calculated is 20.7 au

Granted that is still a very large time step, but this is where the conceptual misstep comes into play. That formula for maximum time step is only half of the picture when it comes to determining the maximum time step to use for a simulation. The other half is the maximum time step for which the integration algorithm is stable. As the documentation states, the Magnus integrator should be stable up to 0.2 au. The formula you give for maximum time step concerns the highest frequency transitions that you will be able to accurately resolve, which is related to signal sampling and Fourier analysis. This maximum time step will only really matter if you are interested in extreme UV/x-ray transitions (wavelength less than about 3 nm).

As far as length of simulation, any length of time will give you information about the entire spectral range up to the maximum frequency dictated by your time step as discussed above. However, your frequency resolution will be 1/T where T is the total time of the simulation, e.g. a simulation length of 1000 au will give you a resolution of 0.027 eV. Always a good idea to checkpoint your simulation so that if you find after analysis that you need better resolution, you can just continue the trajectory.

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Thanks very much. I already noted your point. Sorry for very late reply.

Rangsiman



Quote:Sean Feb 6th 2:12 pm
You have a couple mistakes in your math, as well as a conceptual misstep.

First, the formula for maximum time step is pi over the maximum angular frequency. You have converted from wavelength to period and then have used the period in the formula. The angular frequency corresponding to a period of 1 fs is 2*pi fs^{-1}. So the maximum time step you should have calculated is 20.7 au

Granted that is still a very large time step, but this is where the conceptual misstep comes into play. That formula for maximum time step is only half of the picture when it comes to determining the maximum time step to use for a simulation. The other half is the maximum time step for which the integration algorithm is stable. As the documentation states, the Magnus integrator should be stable up to 0.2 au. The formula you give for maximum time step concerns the highest frequency transitions that you will be able to accurately resolve, which is related to signal sampling and Fourier analysis. This maximum time step will only really matter if you are interested in extreme UV/x-ray transitions (wavelength less than about 3 nm).

As far as length of simulation, any length of time will give you information about the entire spectral range up to the maximum frequency dictated by your time step as discussed above. However, your frequency resolution will be 1/T where T is the total time of the simulation, e.g. a simulation length of 1000 au will give you a resolution of 0.027 eV. Always a good idea to checkpoint your simulation so that if you find after analysis that you need better resolution, you can just continue the trajectory.


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