= Gaussian beam with one transverse coordinate
{c}
{title2=$E=q^{-1/2}e^{-Az^2/q}$}
For the free <parabolic wave equation> $E_x=iE_{zz}/(2k)$, input $E(0,z)=e^{-Az^2}$ with $\operatorname{Re}A>0$ propagates as
$$
E(x,z)=q(x)^{-1/2}e^{-Az^2/q(x)},\qquad q(x)=1+2iAx/k.
$$
The branch is continuous from $q(0)^{-1/2}=1$. <One-dimensional transverse Fresnel propagation> or two <Gaussian integrals> prove the formula. With $A=D^{-2}+ik/(2F)$, the squared envelope is $|q|^{-1}\exp[-2z^2/(D^2|q|^2)]$. The amplitude power is $-1/2$ because there is only one transverse coordinate. A two-transverse-coordinate <Gaussian beam> instead has power $-1$ when both coordinates share the same $q$.
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