# How are the “real” spherical harmonics derived?

How were the real spherical harmonics derived?

The complex spherical harmonics:

$$Y_l^m( \theta, \phi ) = K_l^m P_l^m( \cos{ \theta } ) e^{im\phi}$$

But the "real" spherical harmonics are given on this wiki page as

$$Y_{lm} = \begin{cases} \frac{1}{\sqrt{2}} ( Y_l^m + (-1)^mY_l^{-m} ) & \text{if } m > 0 \\ Y_l^m & \text{if } m = 0 \\ \frac{1}{i \sqrt{2}}( Y_l^{-m} - (-1)^mY_l^m) & \text{if } m < 0 \end{cases}$$

• Note: $Y_{lm}$ is the real spherical harmonic function and $Y_l^m$ is the complex-valued version (defined above)

What's going on here? Why are the real spherical harmonics defined this way and not simply as $\Re{( Y_l^m )}$ ?

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Something to do with Laplace's equation? –  bobobobo May 19 '12 at 22:55
The page actually suggests the answer when it says "The harmonics with $m > 0$ are said to be of cosine type, and those with $m < 0$ of sine type." Recall how one switches between the complex exponential functions $\{e^{imx}\colon m\in \mathbb Z\}$ and the trigonometric functions: it's done with the formulas $$\cos mx=\frac{e^{imx}+e^{-imx}}{2}$$ and $$\sin mx=\frac{e^{imx}-e^{-imx}}{2i}$$ Taking only real parts would not give you the sines.
Since $\cos (-mx)=\cos mx$ and $\sin(-mx)=-\sin mx$, we don't need all values of $m$ in both families. We can remove the redundant functions and enumerate the entire trigonometric basis by $m\in\mathbb Z$ as follows: $\{\cos mx\colon m\ge 0\}\cup \{\sin mx\colon m<0\}$. This is essentially what the wiki page does.
For one thing, it makes sense to keep even and odd basis functions (w.r.t. $\varphi$) separately; for instance if you want to expand an even function in this basis, you only need the even basis functions. But really: if you have a basis for $L^2_{\mathbb C}$ which involves complex exponential functions, and you want a basis for $L^2_{\mathbb R}$, what are you going to do? Of course, you will rewrite exponentials in terms of cosines and sines. –  user31373 May 14 '12 at 20:27