Dear Balamurugan,<br><br>Some answers below and many previous postings:<br><br><div class="gmail_quote">On Thu, Mar 22, 2012 at 4:24 PM, J. K. Balamurugan <span dir="ltr"><<a href="mailto:albertbalagan@gmail.com">albertbalagan@gmail.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Dear WIEN2k developers and users,<div><br></div><div>I have some problem in quantitative understanding the DOS and partial DOS plots/data. I post the question with a case assumption of room temperature (RT) structure of BaTiO3. As it is well known BaTiO3 has tetragonal structure at RT. Now, Ba and Ti has one site and O has two sites. If the DOS & partial DOS are calculated for BaTiO3, I will get total DOS for the unit cell which contains one formula unit of BaTiO3. I will also get partial DOS of Ba, Ti and O1 and O2 (in terms of total and from individual orbital contribution like s, p and d whichever are applicable.) Here are my questions:</div>
<div><br></div><div>1. Will the sum of total DOS individual atoms [i.e., total DOS of (Ba + Ti + O1 + O2)] be exactly equal to total DOS of the unit cell? If the answer is No, why? </div></blockquote><div>No, the interstitial states contribution is missing. Only the area inside the muffin tin is included in the partial DOS. <br>
</div><blockquote class="gmail_quote" style="margin:0pt 0pt 0pt 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div><br></div><div>2. Will the total DOS of any specific atom, say Ti, will be exactly equal to the sum of orbital contributions [i.e., partial DOS of (Ti-s + Ti-p + Ti-d)]? If the answer is No, why? </div>
</blockquote><div>Yes, but I think even further depending on how many l-values are in the basis set. <br></div><blockquote class="gmail_quote" style="margin:0pt 0pt 0pt 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div><br></div><div>Another related question: Let us assume a hypothetical structure where in I have two formula units of BaTiO3 in a unit cell. That is the unit cell has Ba2Ti2O6. Now the question is the following:</div>
<div>3. Should I need to multiply 2 to the total DOS of individual atoms {i.e., 2*[total DOS of (Ba + Ti + O1 + O2)]} to get the total DOS of the unit cell with two formula unit? Will this sum be exactly equal to the actual total DOS of the unit cell which we get as it is from the calculations?</div>
</blockquote><div>Yes. Then no not total DOS of cell, see point 1. <br></div><blockquote class="gmail_quote" style="margin:0pt 0pt 0pt 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div><br></div><div>Please explain me to clear these my questions. I wish also to get into reading any material/user guide/article/document which could help me to have a complete understanding on this issue.</div><div><br>
</div><div>Thanks.</div><div><br></div><div>With kind regards,</div><div>K. Balamurugan.</div><span class="HOEnZb"><font color="#888888"><div><br></div><div><br></div></font></span></blockquote><div>David Tompsett. <br></div>
<blockquote class="gmail_quote" style="margin:0pt 0pt 0pt 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><span class="HOEnZb"><font color="#888888"><div></div><div><br clear="all"><div><br></div>-- <br><font size="4" color="#666600" face="georgia,serif"><i>K. <font>Bala</font>muru<font>gan</font><br>
Pittsburgh, USA.<br><a href="tel:%2B1%20412%20961%205055" value="+14129615055" target="_blank">+1 412 961 5055</a></i></font><br>
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