> On 25 Dec 2022, at 00:26, Md. Jahid Hasan Sagor <[email protected]> wrote: > > > Then why would the calculation be considered computationally large if I > define 10 nm width ( rather I reduce my system size from infinite to 10 nm)? > Can you explain it to me?
Because “reducing the size” (actually, making it finite) would break translational symmetry and thus prevent us from using the Bloch theorem. In an infinite *and periodic* system, Bloch theorem allows us to break an infinite-dimensional Hamiltonian matrix into an infinite number of finite-dimensional matrices (one for every k vector), which can be easily diagonalized one by one. This is not possible in a finite system (or in an infinite and aperiodic one, for that matter). So, surprisingly (or not so surprisingly once you understand the maths) finite systems may be computationally harder than some infinite ones. Stefano B ___ Stefano Baroni, Trieste -- http://stefano.baroni.me _______________________________________________ The Quantum ESPRESSO community stands by the Ukrainian people and expresses its concerns about the devastating effects that the Russian military offensive has on their country and on the free and peaceful scientific, cultural, and economic cooperation amongst peoples _______________________________________________ Quantum ESPRESSO is supported by MaX (www.max-centre.eu) users mailing list [email protected] https://lists.quantum-espresso.org/mailman/listinfo/users
