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In this question I “derive” the Schrödinger equation. A similar argument could be used for the Klein-Gordon’s or Dirac’s equations (with Dirac's equation having one more step of linearization of the Klein-Gordon Hamiltonian).

In short these derivations are as follows:

  1. Admit the plane wave ansatz $$ψ=e^{i(kx-ω t)} \tag{1}$$
  2. Get its derivatives with respect to $x$ and $t$.

$$\begin{equation} \begin{cases} k^2=- ψ^{-1} \frac{∂^2ψ}{∂x^2}\\ ω =iψ^{-1} \frac{∂ψ}{∂t} \end{cases} \tag{2} \end{equation}$$

  1. Make use of the De Broglie’s relations

$$\begin{equation} \begin{cases} E=ħω\\ p=ħk \end{cases} \tag{3} \end{equation}$$

  1. Use the right dispersion relation

$$\begin{equation} \begin{cases} E=\frac{p^2}{2m}+U\\ E^2=c^2p^2+m^2c^4 \end{cases} \tag{4} \end{equation}$$

Substituting (2) into (4) via (3) we get Schrödinger’s and Klein-Gordon’s equations. The thing is, how do we add $U$ into the relativistic one’s if that relation is supposed to be valid for free particles only? (restrict relativity)


**PS: I’m aware that wave equations are not rigorously speaking, derived. They are postulated. Take these “derivations” as arguments in favour for the definition of a certain wave equation.

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  • $\begingroup$ More on potential energy in SR. $\endgroup$
    – Qmechanic
    Commented Apr 10, 2018 at 13:16
  • $\begingroup$ Your equation is wrong, $k^2 = - \partial_{xx} \psi$ should be $k^2 \psi = -\partial_{xx} \psi$, and similarly for the other. $\endgroup$
    – JamalS
    Commented Apr 10, 2018 at 13:17
  • $\begingroup$ @JamalS Thanks. It was missing a $ψ^{-1}$. I already made the changes. $\endgroup$
    – J. Manuel
    Commented Apr 10, 2018 at 14:02
  • $\begingroup$ This question solved my problem. Made a rookie's mistake. Thanks all. physics.stackexchange.com/q/389325 $\endgroup$
    – J. Manuel
    Commented Apr 10, 2018 at 14:38

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