<html><body><div style="color:#000; background-color:#fff; font-family:times new roman, new york, times, serif;font-size:12pt"><div><br></div><div style="color: rgb(0, 0, 0); font-size: 16px; font-family: times new roman,new york,times,serif; background-color: transparent; font-style: normal;">Dear Peter Blaha and users</div><div style="color: rgb(0, 0, 0); font-size: 16px; font-family: times new roman,new york,times,serif; background-color: transparent; font-style: normal;"><br></div><div style="color: rgb(0, 0, 0); font-size: 16px; font-family: times new roman,new york,times,serif; background-color: transparent; font-style: normal;">I have calculated the electronic structure of undoped YBCO and Bi-2212 by MBJ+U where both compound have two CuO2 planes. For YBCO the calculation leads to AFM ground state with a gap of 1.5eV by U=8eV. While the calculations of Bi-2212 reveal metallic properties instead of AFM. I have tried to repeat calculations by U=12
but unfortunately it also leads to metallic properties. By considering U for other atoms such as Bi, Sr, and Ca the calculations again leads to metallic properties. However all calculations have been performed by AFM (i.e. considering two opposite spins for Cu).</div><div style="color: rgb(0, 0, 0); font-size: 16px; font-family: times new roman,new york,times,serif; background-color: transparent; font-style: normal;">What's your advise?</div><div style="color: rgb(0, 0, 0); font-size: 16px; font-family: times new roman,new york,times,serif; background-color: transparent; font-style: normal;">Best Regards</div><div style="color: rgb(0, 0, 0); font-size: 16px; font-family: times new roman,new york,times,serif; background-color: transparent; font-style: normal;">Ali<br> </div><div style="color: rgb(0, 0, 0); font-size: 16px; font-family: times new roman,new york,times,serif; background-color: transparent; font-style: normal;"><br></div><div
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