# How is the Schroedinger equation a wave equation?

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?

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The " in ö is called an umlaut or trema: en.wikipedia.org/wiki/Diaeresis_(diacritic). About the equation being a wave, I have little knowledge of quantum-mechanics, but I can imagine it is connected to the imaginary unit in front of the time derivative –  Michiel Aug 26 '13 at 20:39
Hi user28823, and welcome to Physics Stack Exchange! Since Schroedinger was Austrian, the double-dots would be an umlaut. It's also possible to write that vowel as oe instead of o with double dots, which I changed in your question; hope you don't mind. ;-) –  David Z Aug 26 '13 at 21:12
Both answers were very helpful! thank you –  user28823 Aug 28 '13 at 2:33

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 $e^{ik\xi}$.

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$$A \partial_x^2 u + B \partial_x \partial_y u + C \partial_y^2 u + \text{lower order terms} = 0$$
The wave equation in one dimension you quote is a simple form for a hyperbolic PDE satisfying $B^2 - AC > 0$.
The Schrödinger equation is a parabolic PDE in which we have $B^2 - AC < 0$. It can be mapped to the heat equation.