UT theoretical chemistry code forum

hi,
I am using the dosanalyze.pl script to calculate the d-band center for my system. Setting the weighted average flag to 2.5 (default) gives me 1.6 eV for the d-band center. However, when I change w to 2.0, I get 1.79 eV. Using w=3.0, in turn, gives me 1.52 eV. What is the reasonable value for w? How do I decide on it? Why is w=2.5 the default value. Clearly, my d-band center depends on the choice of w. Can someone please help me understand what w really does and how to choose it correctly. I appreciate your help in advance.
Thanks.

I really don't like this uncertainty associated with the integration limits of the d-band. On one hand, it is dangerous to integrate over the entire DOS because there are high-energy states above the Fermi level which can artificially shift the d-band up, and arbitrarily so if you include more and more bands in the calculation. On the other hand, this automatic window setting also leads to a somewhat arbitrary definition.

I think that the only way the d-band center calculations can be used quantitatively is to pick a fixed integration range (using [e=emin,emax]) for comparison between different systems. This should include all the low energy d-states, and enough states above the Fermi level to include all non-noisy values in the unoccupied d-DOS.

hello,
thanks for your quick reply to my question. Could you please elaborate a bit more about the weighted average flag? What exactly is it doing?
In the code, $cut=0.25*$maxDOS. I am not sure how this is related to the integration limits of the d-band. I would greatly appreciate your help.
Thanks.

It's not a weighted average; it's a width. The idea is to find the width of the distribution, by finding the energy values for which the DOS is 1/4 of the maximum value. The w parameter describes how many such widths to integrate over. So, for example, setting w=2 will integrate over a window twice as large, centered about the energy for which the DOS is a maximum. This is similar to integrating a distribution over 2 standard deviations.

Thank you.