4 ms·
I found this section confusing (emphasis mine): https://physicstravelguide.com/equations/schroedinger_equation#tab__concrete https://physicstravelguide.com/equ
by monktastic1 7y ago
I found this section confusing (emphasis mine):
https://physicstravelguide.com/equations/schroedinger_equation#tab__concrete https://physicstravelguide.com/equations/schroedinger_equati...
"This means that for all systems where the Hamiltonian does not explicitly depend on the time, we known (sic) immediately how the time-dependence of the total wave function Ψ(x,t) looks like (sic), namely: Ψ(x,t) = phi(t)ψ(x) = Ae^(−Et/ℏ)ψ(x).
The only thing we then have to do is to solve the stationary Schrödinger equation Hψ(x) = Eψ(x)."
It sounds like they're saying that all systems with a time-independent Hamiltonian are stationary, which is obviously wrong. This would have confused the hell out of me while learning QM, and dissuades me from learning more on this site.
Have I misread? I can infer what they're trying to say from context and previous knowledge, but that's of little help for someone new.
- contravariant 7y agoYou have misread slightly I fear. Note that he says: >the stationary Schrödinger equation Hψ(x) = Eψ(x)." which is not the original Schrödinger equation, but a Schrödinger equation which is stationary and which forms part of the solution of the original Schrödinger equation.
- monktastic1 7y agoThanks. That's the context I was referring to. Reading the whole thing again twice more, I can still confidently say it would have confused me more than helped me when I was first learning.
- scythe 7y ago>Ψ(x,t) = phi(t)ψ(x) = Ae^(−Et/ℏ)ψ(x) This equation is missing a sum symbol, also it should be e^(-iEt/ℏ), where in the correct case the amplitude does not change. What you get is a sum of products of each occupied stationary state and a time-dependent phase. So the general solution looks like this: https://en.wikipedia.org/w/index.php?title=User:Scythe33/sandbox&oldid=948716529 https://en.wikipedia.org/w/index.php?title=User:Scythe33/san...
- monktastic1 7y agoIndeed. But as another commenter points out, what I think they wanted was to solve only for eigenstates here. Apparently I'm not the only one this confused.