In the real spherical harmonics, where does the sqrt(2) factor come from?

Question

The real spherical harmonics can be written in terms of the complex spherical harmonics:

$$ Y_{\ell m} = \begin{cases} \displaystyle \sqrt{2} \, (-1)^m \, \operatorname{Im}[{Y_\ell^{|m|}}] & \text{if}\ m<0\\ \displaystyle Y_\ell^0 & \text{if}\ m=0\\ \displaystyle \sqrt{2} \, (-1)^m \, \operatorname{Re}[{Y_\ell^m}] & \text{if}\ m>0. \end{cases} $$

Where does the factor of $\sqrt{2}$ come from?

Actually, it's also not clear to me where the $(-1)^m$ comes from either, though I can believe that's related to the Condon-Shortley phase. Is there an intuitive explanation for how the real and complex spherical harmonics are related?

Answer 1

I have seen them written (equivalently) in the form (given by this wiki page) as

$$ Y_{lm} = \begin{cases} \frac{1}{\sqrt{2}} ( Y_l^m + (-1)^mY_l^{-m} ) & \text{if } m > 0 \\ Y_l^m & \text{if } m = 0 \\ \frac{1}{i \sqrt{2}}( Y_l^{-m} - (-1)^mY_l^m) & \text{if } m < 0 \end{cases} $$ Now, Wikipedia page says (quote) "The harmonics with $m > 0$ are said to be of cosine type, and those with $m < 0$ of sine type." The relationship between the complex exponential functions $\{e^{imx}\colon m\in \mathbb Z\}$ and the trigonometric functions being $$\cos mx=\frac{e^{imx}+e^{-imx}}{2}$$ and $$\sin mx=\frac{e^{imx}-e^{-imx}}{2i}$$

It's clear that the relationship between your written version and mine is that I have explicitly expressed the real and imaginary parts.

Answer 2

As for the $(-1)^m$, yes, it depends on the choice of whether using the Condon-Shortley phase or not.

The $\sqrt{2}$ comes from the need of having norm $1$: if you have two unit orthogonal vectors the sum has norm $\sqrt{2}$, in this case $Y^m_l$ and $(Y^m_l)^*=(-1)^m Y^{–m}_l$ are orthogonal and of norm $1$, therefore the real part of $Y^m_l$, which is $$\dfrac{Y^m_l+(-1)^mY^{-m}_l}{2}$$ has norm $1/\sqrt{2}$. To get norm $1$, you multiply by $\sqrt{2}$.