Author Topic: calculate the dipole moment  (Read 5641 times)

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Offline jerry

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calculate the dipole moment
« on: October 4, 2012, 04:21 »
Dear, recently, i have read an article which concentrates on the electronic structure of BN nanoribbons (built by substituting the carbon atoms of GNRs with boron and nitrogen atoms). In that paper they have calculated the transverse dipole moment of BNNR per unite cell, can you give me a script to do this using ATK?

Thanks a lot!

Offline zh

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Re: calculate the dipole moment
« Reply #1 on: October 4, 2012, 07:11 »
From the definition of dipole moment, you may do a bit coding to integrate the charge density  along a specific direction for the calculations of transverse dipole moment.

Offline kstokbro

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Offline jerry

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Re: calculate the dipole moment
« Reply #3 on: October 9, 2012, 03:05 »
Thank you very much for your kind reply!


Offline jerry

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Re: calculate the dipole moment
« Reply #4 on: October 13, 2012, 08:52 »
Dear,

I have refered to the reference manual. But here I want to calculate the transverse dipole moment as the one shown in Table 1 in J. Phys. Chem. C 2009, 113, 21213–21217. Can you give me the script on this which can be applied to calculate the BN nanoribbons?

Thank you very much!

Offline Anders Blom

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Re: calculate the dipole moment
« Reply #5 on: October 13, 2012, 21:39 »
If I understand the article correctly this is just the charge difference of the two edges divided by the width of the ribbon. Thus it's possible to compute from the Mulliken populations, although the precise definition of Dy (in the article) is not crystal clear.

You could also plot the electrostatic potential across the ribbon, either at different positions Z along the periodic cell, or averaged over Z. The slope of this is the electric field, which perhaps is not constant across the ribbon, but probably not rapidly varying either, at least not in the averaged case.
« Last Edit: October 13, 2012, 21:43 by Anders Blom »

Offline jerry

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Re: calculate the dipole moment
« Reply #6 on: October 14, 2012, 03:26 »
Thank you very much for your kind reply :)