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Consider a quantum system described by the wave function $\psi({\bf x}, t)$ and subjected to a time-independent ordinary potential $V({\bf x})$. The relative Schrödinger equation takes the form: $$\underbrace{\left (-\frac {\hbar ^2 \nabla ^2}{2m} + V({\bf x}) \right)}_{\hat H ({\bf x}, \ {\bf p})} \psi({\bf x}, t) = i \hbar \frac \partial {\partial t}\psi({\bf x}, t)$$

Using separation of variables, we write the solution as $\psi({\bf x}, t) = \varphi ({\bf x}) \ \phi(t) $, where:

  • $\phi(t) = \exp(-\frac i \hbar Et)$ solves the equation $ \displaystyle i \hbar \frac d {dt} \phi(t) = E \phi(t)$ , with $E$ constant.
  • $\varphi({\bf x})$ solves the equation $H({\bf x},{\bf p}) \varphi ({\bf x})= E \varphi({\bf x}) $

Question. Is this the most general solution? Or is it just a particular one for a specific guess (a factorized solution)?

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  • $\begingroup$ Does this answer your question ? $\endgroup$ Jul 15, 2021 at 9:29

1 Answer 1

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Consider the following wavefunction:

$$\psi(x,t) = \frac1{\sqrt2}\left(\phi_1(x) e^{-iE_1t/\hbar} + \phi_2(x) e^{-iE_2t/\hbar}\right)\!,$$

Where $\phi_1(x)\,e^{-iE_1t/\hbar}$ and $\phi_2(x)\,e^{-iE_2t/\hbar}$ are stationnary solutions of the Schrödinger equation with $E_1 \neq E_2$. By linearity, $\psi$ is also a solution of the Schrödinger equation and it cannot be factorized.

Now, since $H$ is Hermitian, it can be diagonalized and with some additional work you can show that the factorized solutions that you wrote form a complete basis of the solutions to Schrödinger equation, in the sense that any solution can be written as a linear combination of factorized solutions.

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