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?
