Does anybody have an idea why the integrated DOS is nonzero when DOS is zero? The DOS and Integrated DOS are obtained from DOSCAR. Thx
Hello Hui,
You are referring to the first increase step at -40 in your units?
The overall integration looks quite correct, so: are you sure the DOS is non-zero at -40 or is there a numerical peak that is not resolved in your graph?
Have you looked into the spin-resolved data that separates the DOS contribution from up/down spin-polarization and also integrates them separately?
Maybe you can track it down from resolving this to each ion (which I understand DOSCAR can do for you).
Hope that helps.
Regards.
Hi Hui,
I guess that is a matter of definition of what is DOS and what is integrated DOS. In your DOSCAR file, the density of states (DOS) , is actually determined as the difference of the integrated DOS between two pins, i.e.
n(\epsilon_{i}) = (N(\epsilon_{i}) - N(\epsilon_{i-1})) / \Delta{epsilon}
where Delta{epsilon) is the energy difference in that energy slot (energy difference between two grid point in the DOSCAR file), and N(x) is the integrated DOS:
N(x) = integral between -infinite and x of n(x)
This method conserves the total number of electrons exactly.
Looking at your plot, in the range between -40 and -20, for example, the red line is a constant, so that N(x) and N(x-1) are equal, and their difference, i.e. the DOS, is equal to zero. The plot seems correct to me and fulfills the relation between DOS and IDOS.
Best,
A.
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1 Recommendation
3rd Feb, 2017
Hui Zhang
Lawrence Berkeley National Laboratory
Hi Angelo,
Thank you so much for you explanation. How about the IDOS and DOS around -40?
Regards,
Hui
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12th Feb, 2017
Hui Zhang
Lawrence Berkeley National Laboratory
Hi Christin,
The spin-resolved data is strange. Do you have any idea what is wrong for the DOS? Thank you very much.
-41.913 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-41.743 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-41.574 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-41.404 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-41.234 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-41.064 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-40.894 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-40.724 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-40.554 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-40.384 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-40.215 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-40.045 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-39.875 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-39.705 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00
-39.535 0.0000E+00 0.0000E+00 0.6000E+01 0.6000E+01
-39.365 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-39.195 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-39.026 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-38.856 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-38.686 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-38.516 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-38.346 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-38.176 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-38.006 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-37.836 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-37.667 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-37.497 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-37.327 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-37.157 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-36.987 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-36.817 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-36.647 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-36.478 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
-36.308 0.0000E+00 0.0000E+00 0.2400E+02 0.2400E+02
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13th Feb, 2017
Christin David
Friedrich Schiller University Jena
If I understand correctly, this output shows
energy DOS(up) DOS(dwn) integrated DOS(up) integrated DOS(dwn).
Then everything before -39.705 and after -39.195 seems correct. From this, it seems like a numerical problem, since the program is supposed to obtain the DOS from the integrated DOS.
How about you calculate the DOS yourself from your original data on the integrated DOS, with the equation Angelo pointed out above? Is the peak you'll calculate by hand noticable? What system is this: Do you expect a DOS peak at this energy?
What energies are those, is it pysical to look at such low energy?
Regards.
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16th Feb, 2017
Hui Zhang
Lawrence Berkeley National Laboratory
Hi Christin,
Thank you for your reply. The peak is not negligible. The material is Mg-Al-Y long period ordered structure. Actually, I do not care about the DOS around that low energy. I am just confused about the results and want to find out what is wrong. Again, thank you for your help.
Regards,
Hui
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21st Oct, 2017
Bowen Huang
Hunan University
Hi,
Have you solved this problem? I got a similar problem like yours.
There is no thing on the DOS, but the density has been integrated in IDOS.
Best,
Bowen
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