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?

  • 4
    $\begingroup$ 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 '17 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)? $\endgroup$ – daljit97 Apr 22 '17 at 17:55

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.