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

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...

Mike

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Thanks to whomever made all of the equations look prettier! Now I know how to add equations properly in this forum. –  Mike Williamson Dec 4 '12 at 19:11

It likely relies on the fact that

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

whereas

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

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.

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Hi, yes, I think you are correct. But my question was more fundamental as to WHY this is the case? No definitions of e that I have run across ever use trigonometry to define the number. So how does it "magically" also have this incredibly useful trigonometric property? Doesn't that seem downright crazy to anyone else? –  Mike Williamson Dec 4 '12 at 19:09
@MikeWilliamson The trigonometric link comes from reconciling the power series expansion of $e$ along with the power series expansions of $\sin$ and $\cos$. –  Arkamis Dec 4 '12 at 19:59
Thanks for the "double edit"! The comment regarding the solution for a drum and the connection between ALL complex exponentials and circles is great! I guess my problem was that I was intuitively seeing what you wrote elegantly & explicitly, but not recognizing that the real "magic" is in recognizing that complex exponentials are "magically circular". That's easier for me to get my head around, since it is simply the easiest / simplest way to write a circle as a function: i^x is itself circular, where x is any real number. Thanks so much!! –  Mike Williamson Dec 6 '12 at 0:16

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.

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