5 ms·
As a physical model, as opposed to a purely mathematical entity, don't you need to add operators to the wave function in order to have applicability to observab
by ttctciyf 3y ago
As a physical model, as opposed to a purely mathematical entity, don't you need to add operators to the wave function in order to have applicability to observable phenomena? I'm no physicist, but applying operators to the wave function seems like a computation to me?
- bangkoksbest 3y agoI mean, theoretical physics models are still physics models. Functional analysis is still a valid mathematical tool for physics modeling even if it doesn't (usually) produce finitely calculable solutions. Physicists assign physical interpretation to certain parameters, even if they are not (yet) observable. I suppose there's probably a higher mathematical representation going on here that is finitely computable though. Something like a symbolic representation of the solution functions for the wave function are calculable even if the parameter values are not.