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Re: Raman tensor and raman activity calculation

Alex Granovsky
gran@classic.chem.msu.su


Hi,

I can confirm that Firefly uses the same equations to compute
Raman intensities and activities as those used by you. My guess
is that you do not use the same system of units as Firefly does.
As depolarization ratios are dimensionless, you get the same
values as computed by Firefly regardless of this.

With Firefly, Raman activities has the dimension of Angstrom4/amu
The computed polarizability derivatives components (in Bohr2) are
first multiplied by the squared Bohr to Angstrom conversion factor
(i.e., by 0.52917724924*0.52917724924). Then, they are projected onto
mormal modes (i.e. dotted with normal modes) which are in amu-1/2 units.
Finally, to compute a and g values, one need to square elements of
Raman tensor so the final units are Angstrom4/amu

Please check your units!

Kind regards,
Alex Granovsky





On Sat Aug 18 '12 5:49pm, Siarhei wrote
---------------------------------------
>Hello!
>I try to calculate Raman tensor from
>Firefly output data (using ALPHA POLARIZABILITY
>DERIVATIVE TENSOR and vibrational mode displacements),
>when doing runtyp=RAMAN calculation .
>Then, to check, whether the calculation of Raman tensor
>was correct, or not, I calculate the Raman activity
>and depolarization ratio directly from obtained Raman tensor.
>(r=3g^2/(45a^2+4g^2), r is the depolarization ratio,
>a-alpha, g-gamma - Raman tensor invariants;
>R=45a^2+7g^2, R-raman activity)
>So, the depolarization ratio obtained from
>Raman tensor agree closely with depolarization
>ratio, provided in Firefly output, but Raman activity don't.
>Depolarization ratio agree closely with calculated,
>what means, that gamma and alpha were calculated correctly.
>I did these calculations for different vibrational
>modes, and always depolarization ratio is correct, but
>Raman activity is not.

>Could you please tell how does Firefly calculate Raman activity?
>Or, maybe, Firefly can print Raman tensor in
>output file (there is some group or keyword)?

>Thanks for any answer in advance.


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