Why is the Fermi surface of a metal irregular at 0 K? And how does 'smearing' remedy the situation in DFT calculations?

I'm getting familiarized with DFT and was wondering why the Fermi surface of a metal is irregular at 0 K? It is for the same reason I believe, some 'smearing' needs to be done if the material in question is a metal. How does this smearing change the situation?

• What do you mean by "irregular"? Apr 9 '20 at 3:43
• I think I meant to say 'sharp'. 'Sharp' fermi surface is mentioned in a handful of resources, for example in this textbook: " The Fermi Liquid Model" by Shigeji Fujita & Kei Ito. Apr 9 '20 at 5:55
• The Fermi function has a singularity when $T=0$. This means, physically, that all electrons are below the Fermi energy and there are none above. There are no thermal fluctuations causing the energy region near the Fermi energy to "smear out". The singularity would make the discreet Fourier transform converge slowly at best. Apr 9 '20 at 11:33
• @user1271772 I pretty much figured out the answer. Smearing introduced a fictitious temperature that smoothens out the Fermi surface so that there aren't problems during Brillouin zone integrations in k-space for metals or low-gap materials Jul 23 '20 at 18:31
• @user1271772 yeah I'm sorta active on matter modeling myself. It's a great community and is completely appropriate for these questions on DFT which were earlier difficult to get proper answers on physics stack exchange. Jul 23 '20 at 18:34