3 ms·
This certainly gives me food for thought. Are you saying then that by allowing people to directly observe the wavefunction I'm not really demonstrating uncerta
by sideshowb 10y ago
This certainly gives me food for thought.
Are you saying then that by allowing people to directly observe the wavefunction I'm not really demonstrating uncertainty at all? As that only arises when you force observation via position/momentum?
Also if a massless quantum particle is dispersionless does that mean you can say "the particle is in this small area and will stay here forever"? Doesn't that violate uncertainty?
- leni536 10y agoI would put it this way: Time dependence, therefore dispersion is in direct connection with the model Hamiltonian. The uncertainty principle is independent of time dependence and the model Hamiltonian. I think I made a mistake about massless quantum particles, since in 3D a packet disperses in all directions. But in 1D it works. It can't stay stationary, but it moves with "c" in either direction but it stays in a localized area, still the uncertainty principle still holds.