Author Topic: non magnetic device in LSDA  (Read 4264 times)

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Offline Cyrille

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non magnetic device in LSDA
« on: September 2, 2016, 09:06 »
Dear all

I am facing the following problem: I want to perform a two terminal device calculation of a non magnetic system but with LSDA and starting with a zero magnetization. I know it seems weird since I could just do a LDA (non polarized)  but I have something else in mind:-)
The funny thing is that in the standard DFT cycles (leads and central part) the system remain non-magnetic but suddenly when entering the NEGF cycle the system acquires a magnetic moment as if the NEGF was not starting from the previous DFT.

Could you tell me why is it so?

thanks in advance

Cyrille

Offline Petr Khomyakov

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Re: non magnetic device in LSDA
« Reply #1 on: September 2, 2016, 09:57 »
What kind of system are you studying? It is possible to have local magnetization in non-magnetic materials with defects or some other kind of spacial inhomogeneity. But one should keep in mind that this might also be an artifact of the particular density functional chosen for the DFT calculation.

Could you supply your python script and output log-file?
« Last Edit: September 2, 2016, 09:59 by Petr Khomyakov »

Offline Cyrille

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Re: non magnetic device in LSDA
« Reply #2 on: September 2, 2016, 10:11 »
In fact this a general problem that I already encountered.
My system is an atomic contact. I know it is magnetic but I want to do a non-magnetic calculation within LSDA.
The standard DFT calculation remain non magnetic but as soon as the NEGF cycle start it seems that a "small magnetic kick" is given to the system that switches magnetic.

Cyrille

Offline Petr Khomyakov

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Re: non magnetic device in LSDA
« Reply #3 on: September 2, 2016, 14:12 »
Seeing the actual script and log-file would be of help in understanding the issue. 

Offline Cyrille

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Re: non magnetic device in LSDA
« Reply #4 on: September 2, 2016, 14:43 »
Here is my input script.

Cyrille

Offline Petr Khomyakov

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Re: non magnetic device in LSDA
« Reply #5 on: September 2, 2016, 20:48 »
There are several things that I would suggest to modify in your script.

- The extensions of your device structure appear to be small and need to be enlarged, especially for the left electrode. I enclose an image showing what the structure should look like in a decent setup. In general, there must be a sufficient amount of buffer layers between the actual semi-infinite left (right) electrode and the central region, see http://docs.quantumwise.com/tutorials/atk_transport_calculations/atk_transport_calculations.html?highlight=atk%20transport.

- In the present script, you have the k-point mesh (6,6,10) and (6,6,1) for the left and right electrode, respectively. The k-point mesh should be consistent for the left and right Pd electrodes, e.g.,  set it to  k_point_sampling=(6, 6, 101).  You need a lot of k-points in the transport direction to properly match the Fermi level in the electrodes and central region, see http://docs.quantumwise.com/tutorials/transport_kpoints/transport_kpoints.html?highlight=why%20point .

- The electron temperature might need to be increased to something like  1200 K (which corresponds to broadening of ~0.1 eV) since the system is metallic, see  http://docs.quantumwise.com/tutorials/fe_mgo_fe/fe_mgo_fe.html?highlight=initial%20state

In addition, I would suggest first doing geometry optimization for the corresponding slab structure, allowing Co and some neighboring Pd atoms/layers to relax and setting constraints for all the other atoms, e.g., see the tutotial http://docs.quantumwise.com/tutorials/geometry_optimization/geometry_optimization.html?highlight=geometry%20optimization on rigid and fixed constraint options. This will also allow you optimizing the separation distance between the electrodes.

You will then have a good initial structure for device calculation, and perhaps not even need to do geometry optimization for the device configuration afterwards. But you may do it indeed, if needed.  To optimize the structure, you may just do it with LDA, this will save you the computational time.

     
« Last Edit: September 2, 2016, 20:58 by Petr Khomyakov »