4 ms·
You 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
by contravariant 7y ago
You 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.