Gaussian beam

Remarks

Up to a constant factor, $E = A \, e^{ikz}$ with $A = q^{-1} \exp(ik r^2 / (2q))$ and $q = z - i z_R$. Indeed $1/q = 1/R(z) + i \lambda / (\pi w(z)^2)$, and this $A$ solves the paraxial wave equation. So $w(z)$ is the beam radius at which the amplitude falls to $1/e$ of its value on the axis, and $R(z)$ is the radius of curvature of the wavefronts. At the waist $1/R(0) = 0$, so the wavefront is flat there. Far from the waist, $w(z) \approx \theta z$ with the divergence angle $\theta = \lambda / (\pi w_0)$. The phase $\psi(z)$ is the Gouy phase.

Example: for $\lambda = 633$ nm and $w_0 = 0.5$ mm, $z_R = \pi (0.5 \times 10^{-3})^2 / (633 \times 10^{-9}) \approx 1.24$ m. So the beam stays within a factor $\sqrt 2$ of its waist radius over about $1.24$ m on each side.

Non-example: a uniform plane wave also solves the paraxial wave equation, but its intensity does not decay away from the axis. It carries infinite power, so it is not a Gaussian beam.