The time-independent Schrödinger equation of a free particle in 1 dimension is $$ \begin{equation} -\frac{\hbar^2}{2m}\partial^2_x\psi(x) = E\psi(x) \end{equation} $$ which has solutions in form of $e^{ikx}$, where $k=\frac{\sqrt{2mE}}{\hbar}$ and $E = \frac{\hbar^2k^2}{2m}$.

Any superposition of these functions must be also a solution to the Schrödinger-equation, so let say $$ \psi(x) = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}dk\phi(k)e^{ikx}. $$

Then we expect that $\partial_x^2\psi(x) = - \frac{2mE}{\hbar^2}\psi(x)$ still holds. However, $$ \partial_x^2\psi(x) = \partial_x^2\frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}dk\phi(k)e^{ikx} \\ = \frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}dk \partial_x^2 \phi(k)e^{ikx} \\ = \frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}dk \phi(k)(ik)^2e^{ikx} $$.

So the Schrödinger-equation implies that $$ \frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}dk \phi(k)(ik)^2e^{ikx} = -\frac{2mE}{\hbar^2}\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}dk\phi(k)e^{ikx} $$ That is, $$ \int\limits_{\infty}^{\infty}dk k^2\phi(k)e^{ikx} = \frac{2mE}{\hbar^2}\int\limits_{\infty}^{\infty}dk \phi(k)e^{ikx}, $$ for any functions $\phi(k)$. Does this mean that wave packets do not satisfy the Schrödinger equation?


1 Answer 1


"The time-independent Schrödinger equation[...]" is not the Schrödinger equation, it is a tool used to solve the Schödinger equation. The time-independent Schrödinger equation (TISE) is a result of performing separation of variables on the Schrödinger equation (SE). In this setup you're basically looking for eigenvalues and eigenfunctions of the Hamiltonian operator, with the intention of breaking down the initial conditions in terms of the eigenfunctions to find the time evolution more easily.

The statement, "Any superposition of these functions must be also a solution to the Schrödinger-equation[...]" applies to the SE, not the TISE. You can apply a limited version of it to the TISE, but only to eigenfunctions that have the same energy (eigenvalue).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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