Derivation of the Gaussian Beam Pattern I am currently reading this book by Mireille Levy. It says (on page 40) that the Gaussian beam pattern with half-power beamwidth $\beta$ is gien by:
$$B(\theta)=A\text{exp}\left(-2\text{log}2\frac{\theta^2}{\beta^2}\right)$$
where $A$ is the normalization constant, $\theta$ is the elevation angle and $\beta$ is the half-power beamwidth.
The context is source modelling of the initial vertical field in parabolic equation modelling. I want to understand this equation and understand where it comes from, but I can't find any information online. Could anyone derive it to me or point me to some source, so I can understand it better? Thanks.
 A: The parabolic equation for a beam propagating in the $z$ direction comes from inserting 
$$
\varphi(x,y,z,t) = \psi(x,y,z)e^{i(kz-\omega t)}
$$
into the wave equation 
$$
\frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2}+\frac{\partial^2 \varphi}{\partial z^2}- \frac 1{c^2}\frac{\partial^2 \varphi}{\partial t^2}=0
$$
to get 
$$
 \frac{\partial^2 \psi}{\partial z^2}+2ik \frac{\partial \psi}{\partial z}- k^2 \psi +  \frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+ \frac{\omega^2}{c^2}\psi.
$$ We now drop   the ${\partial^2 \psi}/{\partial z^2}$ term on the grounds that we have included all the fast changes in $z$    in the exponential factor $e^{ikz}$. The result is the paraxial wave equation
$$
2ik \frac{\partial \psi}{\partial z}=- \left(\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}\right) +\left(k^2-\frac{\omega^2}{c^2} \right)\psi. 
$$
This looks like the time-dependent Schroedinger equation, but with time replaced by the distance $z$ down the beam and Planck's constant $\hbar$  by twice the wavenumber $2k$. We can therefore plug in the textbook  gaussian solution to Schroedinger to find what the optical beam does. 
There is a good account of this on Wikipedia: https://en.wikipedia.org/wiki/Gaussian_beam
