QuantumATK Forum
QuantumATK => General Questions and Answers => Topic started by: Dipankar Saha on July 1, 2015, 22:04
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1)What are these....Local Struct. Analysis, Coordination no. analysis....and Centrosymmetry Analysis??
It will also be of great help.......if you can comment on___
a) Projected Local DOS ....
Actually in case of a device..., how the method_' LDOS' differs from the 'DDOS'??
b) Partial charge with classical potential....
What does it signify....when all of the atoms in a system...show the '0' corresponding partial charge (e) values ...?!!
Regards_
Dipankar
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1) see reference manual http://www.quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/ref.localstructure.html
read also notes at the bottom.
2) LDOS and DDOS methods correspond to LocalDOS (http://www.quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/ref.localdevicedensityofstates.html) and DeviceDOS (http://www.quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/ref.devicedensityofstates.html)
3) see http://quantumwise.com/forum/index.php?topic=3485.msg15933#msg15933
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Thank you Umberto...!!!! :)
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1)
Looking at the Constructor for the Device Density of States object,
I find that the default option for the "contributions" is "All" ....Thus,
the DDOS computed...considering the spatial Density matrices...for both the Left and Right electrodes;
makes it different from the bulk one..... Is this the key point ??
2)
Again...as it was mentioned....
DDOS...can be obtained by integrating LDOS...over all space and energy.../ That's why we specify a
energy range in case of the DDOS...whereas, it's a single energy value for LDOS...!!
For LDOS....can we change the default energy of 0 eV....to any other value??
3)
The thread you mentioned...doesn't portray the details...that under what cond.__
the atoms in any system... may show the '0' corresponding partial charge (e) values ??!!
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(3) The PartialCharges are zero if the classical potential you selected does not use coulomb interactions to model the electrostatics of the system. In this case no partial charges on the atoms are defined and that is what the PartialCharges analysis object tells you.
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Okayy... !!! That's the reason.....
Thanks Julian for your reply... :)
Regards_
Dipankar
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1) yes, in the notes is explained that DDOS is computed via the spectral density matrix of left and right contributions.
withr reference to http://www.quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/chap.negf.html#sect2.negf.greenfunction
2) yes of course, for LDOS you can specify an energy at which it is calculated by using the energy parameter. See the same notes.
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Thanks for your reply Umberto...!!! :)
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Besides, I wanted to know one more thing.....
May be previously also ...I asked a similar question....
But, let be more specific this time, .....with a very simple example..... :)
For the purpose.... I take the cut-plane images showing n(r) and del_n(r) ...for the unit-cell moS2 !!! (Please find the attachment....)
1) If n (r) is the final picture....(i.e., something at the end of the self consistent calculation)...then charge associated with all the Mo and S atoms are positive....!!! what does it actually mean??
2) Again, del_n(r) for the Mo atom....... shows a much "-Ve" value..../ What does that signify??!!
Best_
Dipankar
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An electron density of valence electrons can only be positive, hope it is clear.
Please see the ElectronDensity and ElectronDifferenceDensity entries in the reference manual, scroll down until notes.
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Thank u... Umberto for your reply....!!! :)
Please see the ElectronDensity and ElectronDifferenceDensity entries in the reference manual, scroll down until notes.
It's not about.... what that is written there in the reference manual...!!! (In fact, that's the sole purpose of bringing a simple unit cell into the picture..... :D)
Is it possible.....to tell that.....;
why del_n(r) shows a much "-Ve" value for 'Mo'....while reaching the final n (r) state at the end of the self consistent loop...??!!
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I really don't see the purpose of these questions, as the answer is obvious. The value of the electron difference density is negative on Mo because there is a net charge transfer of electrons between the atoms. In your results, there are less electrons around the Mo atom in the self-consistent state (when the atoms are interacting) compared to an isolated Mo atom.
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Thanks a lot Anders for the reply...!! :)
Yeah...here it comes.... as you say.... ,
In your results, there are less electrons around the Mo atom in the self-consistent state (when the atoms are interacting) compared to an isolated Mo atom.
Is it the difference, at the end of the self-consistent loop.... (meaning, {elec_density_for an atom, considering the interactions } - {elec_density_isolated atom})
Or,
a self-consistent state, shows.....what is the net charge on Mo atom after the redistribution ??
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It will be of great help....if you can please answer.... :)
Regards_
Dipankar
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The density is not "for an atom", it's the real-space electron density. And as the manual says (see http://quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/chap.atkdft.html) the difference density is the difference in density in the SCF state compared to the neutral atom case.
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Many thanks Anders for your help....!!! :)
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If the ElectrostaticDifferencePotential.... relates to Hartree Potential only.....;
then how can I find a similar change in [Hartree Potential + ionic contributions]... for the diff. in density??
Thanks & Regards_
Dipankar
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The ionic contribution is independent of the density
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Dr. Anders Blom,
Perhaps...that might be for the pseudo potential term... (though I'm not sure about the non-local part :-\) .../
But, what if we consider the other term, that is the ...... electro-static potential due an external filed, calculated through the Poisson solver... ......?? Is this calculated considering zero electron density?? If so...then, should it be added to the Electostatic Diff. Potential ??
Thanks in advance..... :)
Regards_
Dipankar
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I'm not really sure what your actual question is. All terms are described here: http://quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/chap.atkdft.html#sect2.atkdft.effectivepotential
The only trick thing is that we don't actually compute the Hartree potential, but only the difference compared to the neutral atom system. That's it.
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The only trick thing is that we don't actually compute the Hartree potential, but only the difference compared to the neutral atom system. That's it.
Okayyyy..., now I understand.... / Thanks again Anders.... :)