# relation between Schrodinger equation and wave equation [duplicate]

I have always been confused by the relationship between the Schrödinger equation and the wave equation.

$$i\hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2+ U \psi \hspace{0.25in}\text{-vs-}\hspace{0.25in}\nabla^2 E = \frac{1}{c^2}\frac{\partial^2 E}{\partial^2 t}$$

Because of the first derivative, the Schrödinger equation looks more like the heat equation.

Some derivations of the Schrodinger equation start from wave-particle duality for light and argue that matter should also exhibit this phenomenon.

In some notes by Fermi, it was derived by comparing the Fermat least time principle $\delta \int n \;ds = 0$ and Maupertuis least action principle $\delta \int 2T(t) \; dt = 0$.

Was this ever clarified? How can we see the idea of a matter-wave more quantitatively?

To summarize, I am trying to understand why the Electromagnetic wave equation is hyperbolic while the Schrodinger equation is parabolic.

## marked as duplicate by Brandon Enright, akhmeteli, John Rennie, user10851, jinaweeDec 25 '13 at 14:34

• For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. – Qmechanic Jul 27 '14 at 17:50

To see the idea of a matter-wave more quantitatively and to relate it to the electrocmagnetic fields, it can be helpful to consider a beam propagating in one dimension. For example, a laser (or other coherent light source) propagating down some path. In this scenario, you can treat the transverse laplacian and the z-derivatives (z being the direction of propagation) independently.

The wave equation for EM waves is:

$$\hspace{0.25in}\nabla^2 E = \frac{1}{c^2}\frac{\partial^2 E}{\partial^2 t}$$

this differential equation has plane wave solutions, so without loss of generality you can can examine a field of the form:

$$E(\vec{r}) = A(\vec{r}) e^{i\omega t+ik_0t}$$

$$\nabla^2 \vec{E} = \partial_{xx} \vec{E} + \partial_{yy}\vec{E} + \partial_{zz}\vec{E}$$

$$\nabla^2 \vec{E}= e^{i\omega t + ik_0 z} \partial_{xx} A(\vec{r}) + e^{i\omega t + ik_0 z} \partial_{yy} A(\vec{r}) + \partial_{zz}A(\vec{r})e^{i\omega t + ik_0z}$$

$$e^{i\omega t +ik_0z} \nabla^2 A(\vec{r}) + 2ik_0 e^{i\omega t +ik_0z}\partial_z A(\vec{r} ) - k_0^2 e^{i\omega t +ik_0z} A(\vec{r})$$

The wave equation can then be written as: $$e^{i\omega t +ik_0z}\nabla^2 A(\vec{r}) + 2ik_0 e^{i\omega t +ik_0z} \partial_z A(\vec{r} ) - k_0^2 e^{i\omega t +ik_0z} A(\vec{r}) + \frac{\omega^2 n^2}{c^2}A(\vec{r}) e^{i\omega t +ik_0z} = 0$$ By dropping the common exponent, this can be simplified to: $$\nabla^2 A(\vec{r}) + 2ik_0 \partial_z A(\vec{r} ) - k_0^2 A(\vec{r}) + \frac{\omega^2 n^2}{c^2}A(\vec{r}) = 0$$

And define the transverse laplacian to be

$$\nabla^2_\perp = \nabla^2 - \frac{\partial^2}{\partial z^2}$$

And by making the slowly varying envelope approximation:

$$\frac{\partial^2}{\partial z^2}A(\vec{r}) \ll k_0 \frac{\partial}{\partial z}A(\vec{r}) \ll k_0^2A(\vec{r})$$

Also use the dispersion relation: $$k_0 = \frac{n\omega}{c}$$

The wave equation can then be written as:

$$i\frac{\partial A(\vec{r})}{\partial z} = \frac{1}{2k_0}\nabla_\perp^2 A(\vec{r})$$ which is in the same form as the Schroedinger equation in the absence of an external potential with the exception that instead of varying with time $t$, it varies with position along the axial direction, $z$.