Wave equations take the form:
$$\frac{ \partial^2 f} {\partial t^2} = c^2 \nabla ^2f$$
But the Schroedinger equation takes the form:
$$i \hbar \frac{ \partial f} {\partial t} = – \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$
The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation? And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions?
Best Answer
Actually, a wave equation is any equation that admits wave-like solutions, which take the form $f(\vec{x} \pm \vec{v}t)$. The equation $\frac{\partial^2 f}{\partial t^2} = c^2\nabla^2 f$, despite being called "the wave equation," is not the only equation that does this.
If you plug the wave solution into the Schroedinger equation for constant potential, using $\xi = x - vt$
$$\begin{align} -i\hbar\frac{\partial}{\partial t}f(\xi) &= \biggl(-\frac{\hbar^2}{2m}\nabla^2 + U\biggr) f(\xi) \\ i\hbar vf'(\xi) &= -\frac{\hbar^2}{2m}f''(\xi) + Uf(\xi) \\ \end{align}$$
This clearly depends only on $\xi$, not $x$ or $t$ individually, which shows that you can find wave-like solutions. They wind up looking like $e^{ik\xi}$.