Dear Theo,
How do you model the paramagnetic state? Please note that "paramagnetic" and "nonmagnetic" state is not the same thing. For nonmagnetic calculations there is a factor of 2 due to spin degeneracy when computing sums over electronic states (i.e. in charge density, occupation matrix, and other quantities). If you take a nonmagnetic material (e.g. LiCoO2) and model it as a spin-polarized system (nspin=2), the value of U will be the same as when modeling it as nonmagetic (because the magnetization will be zero). So I do not understand why do you have a factor of 2 difference for U in your simulations. Greetings, Iurii -- Dr. Iurii TIMROV Senior Research Scientist Theory and Simulation of Materials (THEOS) Swiss Federal Institute of Technology Lausanne (EPFL) CH-1015 Lausanne, Switzerland +41 21 69 34 881 http://people.epfl.ch/265334 ________________________________ From: users <users-boun...@lists.quantum-espresso.org> on behalf of Theo Weinberger <ti...@cam.ac.uk> Sent: Tuesday, September 12, 2023 11:21:03 AM To: users@lists.quantum-espresso.org Subject: [QE-users] Spin counting in hp.x with paramagnetic metals Dear Quantum Espresso Users, I have been using the hp.x code to calculate the Hubbard-U corrections for correlated metallic systems in both their spin-polarised and paramagnetic states. In several materials I have noticed that the Hubbard-U value determined for a material in its paramagnetic ground state is approximately twice that compared to when a spin-polarised ground state is assumed (with all other parameters kept the same). I was wondering whether anyone had any insight into how the accounting for spins in occupied Hubbard states works for the hp.x code and whether the paramagnetic implementation of hp.x perhaps counts all spins states twice resulting in this larger value. Thank you in advance, Theo Weinberger PhD Student University of Cambridge
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