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 $A e^{i(kx-\omega t)}$?

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    $\begingroup$ @AccidentalFourierTransform or to say it less hard: The typical "derivation" is basically only a motivation how the S eq can be reached by starting from classical mechanics. By the way, when we start with the Liouville's Theorems as an Axiom, shouldn't we be able to actually derive the Schrödinger erquation? $\endgroup$
    – user_na
    Jul 24, 2018 at 17:23
  • $\begingroup$ @user_na No, Liouville's theorem does not imply Schroedinger's equation. $\endgroup$ Jul 24, 2018 at 18:35
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    $\begingroup$ @AccidentalFourierTransform that comment is not helpful at all. OP is looking at a heuristic derivation of the Schrodinger equation and has asked a very good question about it. OP has even zeroed in on the root of the problem all by themself. Such behavior should be encouraged here. This question really does invite interesting and useful answers on how to think about constructing solutions to the Schrodinger equation from linear superpositions of solutions that obey a heuristic physical idea. $\endgroup$
    – DanielSank
    Jul 24, 2018 at 18:51
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    $\begingroup$ @DanielSank We'll have to agree to disagree. What is not helpful is to encourage the attitude that this "heuristic derivation" useful. It is not. There is nothing else to say. The only correct answer to OP's question is that they should forget about this "derivation" altogether; that way, all problems go away. The "linear superposition" idea is not helpful either, because it only works for free problems, which do not represent real life situations at all. Developing intuition based on unrealistic problems is very dangerous. $\endgroup$ Jul 24, 2018 at 19:40
  • $\begingroup$ For extended discussions on "derivations" of the Schrödinger equation, see physics.stackexchange.com/q/142169/50583 and its linked questions. $\endgroup$
    – ACuriousMind
    Jul 24, 2018 at 19:45

2 Answers 2


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.

  • $\begingroup$ OP put the word "derivation" in quotes. I think this was done to indicate that they know there is no true derivation. Harping on that point seems to miss the spirit of the question, which is about how heuristically constructing an eigenvalue equation in its own diagonal basis relates to that equation and its solutions in a non-diagonal basis. $\endgroup$
    – DanielSank
    Jul 25, 2018 at 1:11

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.


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