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Author Topic: Is the gate added correctly?  (Read 1031 times)
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hol28
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« on: August 28, 2012, 16:27 »

Hi everyone,
I'm now considering studying the conductance of a device with H+ involved. The geometry is included in the file Pt.py. In order to simulate the H+, I add a gate just on the bottom of the atom H. By adjusting the gate, the atom H can be turned into H+. But I'm not sure whether it is reasonable using this method. In addition, what boundary condition should I use in this case?
Any suggestion would be appreciated. Thank you very much!
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Anders Blom
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« Reply #1 on: August 28, 2012, 17:07 »

That's probably a bit too violent Smiley I would suggest adding an onsite shift on the H atom instead (see http://www.quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/ref.atomicshift.html).
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hol28
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« Reply #2 on: August 28, 2012, 23:38 »

Hi, I have tried your suggestion, but another thing puzzled me. When I added an onsite shift of +15*eV on H atom, the  Mulliken Population Report for H is 1.745; when onsite shift is -15*eV, the value is 2.327. It seems that the charge of H should be larger than ~1.00 for one case and smaller than ~1.00 for another. Could you please give me some suggestions for this problem? Thank you so much!
That's probably a bit too violent Smiley I would suggest adding an onsite shift on the H atom instead (see http://www.quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/ref.atomicshift.html).
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Anders Blom
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« Reply #3 on: August 29, 2012, 08:59 »

You cannot instruct the H atom to be in a particular charge state, since the Mulliken population of each atom will be determined by the self-consistent state of the whole system. Thus the relevant quantity to compare to is not the isolated H atom, but the self-consistent state of the H atom in the given atomic environment. I am sure (or at least I very much hope) that if you compare the 3 calculations with +15 eV, 0 eV, and -15 eV, you will see a net additional positive/negative charge on the H atom in the two cases with a finite shift, compared to no shift.
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hol28
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« Reply #4 on: August 29, 2012, 16:56 »

Hi, thank you very much! Yes, I think the result should be like what you say, but what I get is weird. I have read the document you mentioned to add the site shift, but since the case is a little different, I'm not sure whether it is right in my case. Could you please help me check if the method I use to add the site shift is correct? 
Thank you so much!
You cannot instruct the H atom to be in a particular charge state, since the Mulliken population of each atom will be determined by the self-consistent state of the whole system. Thus the relevant quantity to compare to is not the isolated H atom, but the self-consistent state of the H atom in the given atomic environment. I am sure (or at least I very much hope) that if you compare the 3 calculations with +15 eV, 0 eV, and -15 eV, you will see a net additional positive/negative charge on the H atom in the two cases with a finite shift, compared to no shift.

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Anders Blom
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« Reply #5 on: August 29, 2012, 22:43 »

It looks correct. But I think a shift of 15 eV is also not suitable, it's probably way too much, which may explain the strange results.
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hol28
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« Reply #6 on: September 4, 2012, 16:52 »

Hi, I have tried other site shifts, but the results are still very strange. When no site shift is added, the mulliken charge for H is 0.903. When the site shift is +10, the value is 1.463; when -10, it is 1.751. +8: 1.299; -8: 1.257. +5:1.095;-5:0.951. +3:0.994;-3:0.903. The charge of H atom is always increased. Could you please give me other suggestions? Thank you so much!
It looks correct. But I think a shift of 15 eV is also not suitable, it's probably way too much, which may explain the strange results.
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« Reply #7 on: September 6, 2012, 15:34 »

It's a bit hard to tell, the system is after all rather artificial, but I guess it means you can't really expect to have a H+ ion in this place, in this system.
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hol28
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« Reply #8 on: September 6, 2012, 16:05 »

OK. Thank you all the same!
It's a bit hard to tell, the system is after all rather artificial, but I guess it means you can't really expect to have a H+ ion in this place, in this system.
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