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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, jinawee Dec 25 '13 at 14:34

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  • $\begingroup$ 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. $\endgroup$ – 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, $z$.


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