"Fun" question: anyone know why $e$ (Euler's Number) was chosen for wave functions?

Question

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

Answer 1

It likely relies on the fact that

$$e^{ix} = \cos{x} + i\sin{x}$$

whereas

$$2^{ix} = \cos{(\ln{(2)} x)} + i\sin{(\ln{(2)}x)}$$

which leads to ugly formulas.

EDIT: This is great resource, by John Cook, on why $e$ is used: http://www.johndcook.com/blog/2012/11/15/logarithms/ It deals with logarithms, but all arguments hold here too.

DOUBLE EDIT: I'll elaborate a little be more. The Laplacian operator is defined by $\Delta f = \bigtriangledown \bigtriangledown f$, and in the 1d case it is simply the second derivative. The eigenfunctions and eigenvalues for the Laplacian are defined as the solution to

$$ \Delta f = \lambda f $$ This is also the mathematical definition of a drum, which has lots of circular and sinusoidal properties. One particular solution (for the 1d case) is $\lambda = 1, f(x) = e^{ix}$. But, in general, the solution is

$$e^{\sqrt{\lambda}\;ix}$$

which can be molded into any exponential. For example, setting $\lambda = (\log{2})^2$ gives us $2^{ix}$. So what does this mean?

It means there is nothing special about $e$!! The real connection is between all complex exponentials and circles. We merely choose $e$ for mathematical convinence.

Answer 2

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.