I know that the general one-dimensional Schroedinger equation is given by: $$-\frac{\hbar^2}{2m} \frac{\partial^2\Psi(x,t)}{\partial x^2} + U(x)\Psi(x,t) = i\hbar \frac{\partial \Psi(x,t)}{\partial t} $$

The source I am using mentions: "If the potential energy function is nonzero, these sinusoidal waves do not satisfy the Schroedinger equation". Why? I thought the reason why the general equation included $U(x)$ was exactly to address this problem. Then it continues by: "However, we can still write the wave function for state of definite energy $E$ in the following form" $$\Psi(x,t) = \psi(x)e^{iEt/\hbar} $$ I understand that this comes from $$ \Psi(x,t) = Ae^{i(kx - wt)} $$ But why a state of definite energy must have potential energy non-zero?

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    $\begingroup$ I'm not sure what exactly you're asking - I think you're misunderstanding what your source is trying to say: If $U(x)\neq 0$, then $\psi(x) = \mathrm{e}^{\mathrm{i}(kx-\omega t)}$ is not a solution of the Schrödinger equation anymore. What is the question about that? $\endgroup$
    – ACuriousMind
    Apr 11, 2017 at 11:18
  • $\begingroup$ @ACuriousMind why is it not a solution to the Schrödinger equation anymore? So why does the general one-dimensional Schrödinger equation includes U(x)? $\endgroup$
    – user63248
    Apr 22, 2017 at 17:55

1 Answer 1


The form $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$ does not "come" from $Ae^{i(kx-\omega t)}$ but thee from basic ansatz $\Psi(x,t)=\psi(x)\Phi(t)$ used to solve partial differential equations.

The form $\Psi(x,t)=\psi(x)\Phi(t)$ is used to convert the partial differential equation to a pair of ordinary differential equations connected by a separation constant.

Inserting $\Psi(x,t)=\psi(x)\Phi(t)$ into the time-dependent Schrodinger equation produces a result easily rearranged to \begin{align} \frac{1}{\psi(x)}\left(-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+U(x)\psi(x)\right)=\frac{1}{\Phi(t)}\left(i\hbar \frac{d\Phi(t)}{dt}\right)=E \end{align} with $E$ the separation constant (to be identified with the energy). It is possible to solve the $\Phi$ equation immediately as it is independent of the potential: $$ \Phi(t)=e^{-iEt/\hbar}\, , $$ from which $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$ follows, with $\psi(x)$ a solution to the time-independent Schrodinger equation $$ -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+U(x)\psi(x)=E\psi(x)\, . \tag{1} $$ The form $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$ thus holds for arbitrary potential $U(x)$ provided $\psi(x)$ satisfies (1).


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