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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
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2 Answers

up vote 11 down vote accepted

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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Both are types of wave equations because the solutions behave as you expect for "waves". However, mathematically speaking they are partial differential equations (PDE) which are not of the same type (so you expect that the class of solutions, given some boundary conditions, will present different behaviour). The constraints on the eigenvalues of the linear operator are also particular to each of the types of PDE. Generally, a second order partial differential equation in two variables can be written as

$$A \partial_x^2 u + B \partial_x \partial_y u + C \partial_y^2 u + \text{lower order terms} = 0 $$

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.

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