2 ms·
Let's say you want to simulate three carbon atoms with near perfect accuracy (not considering constituent quarks, radiative effects, or relativistic effects).
by Xcelerate 12y ago
Let's say you want to simulate three carbon atoms with near perfect accuracy (not considering constituent quarks, radiative effects, or relativistic effects). We'll remove the BO approximation though. So that's 2^(3 atoms * 3 coordinates * 3*12 particles) = 2^324. That's more than the number of particles in the universe. A classical computer that can handle that is simply not gonna happen. And that's just three atoms.
- ScottBurson 12y agoAre you sure it shouldn't be 2^(3 + 3 + 3*12) = 2^42 ? That would still be a pretty big number, but not yet astronomical.
- ballstothewalls 12y agoI understand that is massive, but aren't computers today doing things that seemed utterly impossible forty years ago?