Does the Schroedinger Equation Make Sense?

In the Schrodinger equation: how can they set equal to each other: (dy/dx)(dy/dx) = d(dy/dx)/dx : first derivative squared = second derivative ? Kinetic energy = 1/2 m v.v = 1/2 m.m v.v /m = (mv)(mv) / 2m = p.p /2m In order to beget an oscillator of some kind, a lot of books say: p.p = second derivative. Example: f(x) = x.x.x dy/dx = 3 x.x (dy/dx)(dy/dx) = 9 x.x.x.x d(dy/dx)/dx = 6 x
- Richard Kroot (age 53)
Rotterdam, Holland

That's a very nice question, one that occurs to many students as we first try to grasp the Schrödinger equation. Somehow it seems at first like we should have a genuine momentum p that depends on ψ, and then get p² by squaring that, ψ and all. That's not how it works, though.  

The equation is strictly linear, which means that if you multiply ψ by anything, say 2, you still get a solution to the same equation. That only works if each term in the equation has a plain ψ in it, no ψ² or ψ³ or anything like that. So what's squared to get p² is the operator which does something to ψ, not the thing that you get after it operates on ψ. It's the second derivative that's needed, not the square of the first derivative. As you say, they can't be the same thing. 

Mike W.

posted without vetting until Lee returns from the Serengeti 

(published on 10/22/2014)