Assumption in the derivation of Schrödinger's equation

Question

I read on the "derivation" (with some assumptions) of the Schrödinger equation. The idea is to start from $$T + U = E $$ wher $T$ is kinetic energy, $U$ is potential energy, and $E$ is total energy. Then, we assume that the wave function has the form $\Psi = Ae^{i(kx-\omega t)}$ and go from there to get the Schrodinger equation. However, I have read examples where the resultant $\Psi$ solved from the Schrodinger equation is not of the form $A e^{i(kx-\omega t)}$ (e.g. free particle in a potential well). If we derive the Schrodinger equation by assuming $\Psi = A e^{i(kx-\omega t)}$, but the solutions that come out from the Schrodinger equation are not of that form, then aren't we sort of contradicting ourselves?

My guess: Does this have something to do with linearity in quantum mechanics as a whole perhaps, i.e. expressing $\Psi$ as a sum of $A e^{i(kx-\omega t)}$?

Answer 1

It is extremely misleading to call the argument you referred to as a "derivation" of the Schrödinger equation $-$ to put it simply, it is no such thing, and there is no such thing. The Schrödinger equation is one of the fundamental postulates of quantum mechanics and we know of no deeper justification for it.

The argument you've sumarized is best described as a justification of why the Schrödinger equation is, hopefully, not too much of an unreasonable thing to try. It is an argument that makes a bunch of leaps of logic, some of which you've already identified, and it cannot be fixed. That doesn't make the argument any less useful, but to the degree that you find logical flaws in it, they are mostly really there.

Answer 2

There is no derivation of Shroedinger equation (Seq), since it is a postulate (axiom) of quantum mechanics. These arguments you used are best thought as “motivations” or even simplistic arguments to make that axiom self-evident.

There are more assumptions on that “derivation” than you may be aware of. They rely on the ideas that the simplest wave form as we know, is the harmonic wave ($\psi = Ae^{i(kx-\omega t)}$), and the simplest state of motion is the motion of a free particle ($U=const$). Therefore, what you are admitting above is only that by connecting these concepts, if the wave equation is “well behaved” than the wave form of a free particle should be harmonic. This would be true just for this particular case, and no other.

Those assumptions don’t say anything on what would be the form of $\psi(x,t)$ under a different setup, or what is the form of $\psi$ when the force acting on the particle is not constant ($U=f(x,t)$). At this point, all one can do is to promote the S eq found for the free particle as invariant under the action of any force, and then check if the experiment much the expectations.

Now you can see that if $U=\frac{1}{x}$ the form of $\psi$ will be different than the harmonic wave.

PS: By the way, it is easy to check that the harmonic wave is a solution for the S eq when the potential is constant.