1

In this spherical harmonics paper, there's a periodic function $\Phi(\phi)$ defined as

$$\Phi(\phi)= \bigg\{ \begin{array} \\e^{im\phi} \\e^{-im\phi} \end{array} \space m = 0,1,2,3...$$

Is $e^{im\phi}$ the same thing as the more conventional $e^{ix}$?

mavavilj
  • 7,270
  • Well it's not quite the same because $\phi \in [0,2 \pi)$ is an azimuthal angle where as $x$ usually denotes the real line. But sure, it's still a complex exponential: $\exp(i m \phi) = \cos(m \phi) + i \sin(m \phi)$. – okrzysik Jun 28 '16 at 09:16
  • Is that $im(\phi)$ as in $i\cdot m\cdot \phi$ or as in "the imaginary part of $\phi$"? – 5xum Jun 28 '16 at 09:17
  • @5xum Good point. I had read it the imaginary part, but since $m$ is a variable in $\Phi$, then it must be $i \cdot m \cdot \phi$. – mavavilj Jun 28 '16 at 09:18
  • @mavavilj if you look in the document it's not written as the imaginary part as you've written in your post, also in the definition is states that $m$ is an integer. – okrzysik Jun 28 '16 at 09:20

1 Answers1

1

Answer: $e^{i m \phi}$ is the same thing as $e^{i x}$ when $x = m \phi$. Note that $i m \phi$ stands for the product $i \, m \, \phi$ and not the imaginary part of $\phi$ here.

This should be interpreted as a family of periodic functions $e^{i m \phi}$ and $e^{-i m \phi}$ indexed by all non-negative integers $m = 0, 1, 2, \dots$.

Svinto
  • 829
  • Ok, now it addresses the actual question as well as a small clarification. From the paper it is quite clear that it is the product that is meant. – Svinto Jun 28 '16 at 09:25