TITLE: "Fun" question: anyone know why $e$ (Euler's Number) was chosen for wave functions?
QUESTION [8 upvotes]: First, let me say that this is merely something I have always wondered about, and can never seem to find a good reference for.  I simply want to know... the geek in me.

Why was $e$ (Euler's Number) chosen for wave function descriptions?  For instance:

$$\Phi(x, t) = Ae^{i(kx - \omega t)}$$
It's really the $i$ that's doing the work of making a circular form here, while the $e$ is simply making the scaling more friendly to what we're used to.  For instance, let's compare $2^{ix}$ versus $e^{ix}$.  When $x=0$, they are both 1.  To get them both to reach $i$, $x = \pi / 2$ for $e^{ix}$ and $x \approx 2.26618$ for $2^{ix}$.  Similarly, for all the other quadrants of the circle, an equivalent factor can be found for $2^{ix}$, scaling linearly, of course.

So the scaling might look a bit less "pretty", but it is completely functional using $2^{ix}$ instead of $e^{ix}$.

So, I guess my question is twofold:


Why is the wave equation using $e$, other than because it supplies the "proper" scaling factor to make it friendlier with circular equations?  (I.e., $2 \pi = 0$, brings us back to where we started.)
What is it about $e$ that makes the scaling work out?  Euler's number was derived from $ \lim \ (1 + 1/n)^n$ , which doesn't, to me, suggest anything particularly circular to it.  (In fact, from that definition, it also doesn't immediately suggest why it's derivative is equal to itself, either, but that's a different question for another day!)  Just seems awful serendipitous to me, too much so, which makes me suspect a connection I don't know about...
Thanks in advance!

 Mike

REPLY [5 votes]: By using a base of $e$, derivatives are simpler. As I think you recognize, $\frac{d}{dx}e^{ikx}=ik\,e^{ikx}$. Because of this, when initial value $v_0$ and initial relative slope $r_0$ are known, they fit right in: $v_0\,e^{ir_0x}$. All of this is uglier with a number other than $e$ for the base.