= Solution
Use the time convention $e^{-i\omega t}$. The free-space <Helmholtz equation>, with $\psi=e^{ikx}E$, gives the exact reduced equation
$$
E_{xx}+2ikE_x+E_{zz}=0.
$$
For a forward <plane wave> at angle $\alpha$, $E_x=O(k\alpha^2E)$ and $E_{zz}=O(k^2\alpha^2E)$, whereas $E_{xx}=O(k^2\alpha^4E)$. Thus the <paraxial approximation> neglects $E_{xx}$ relative to $2ikE_x$. \b[The reduced field satisfies the <parabolic wave equation>]
$$
\boxed{2ikE_x+E_{zz}=0,\qquad E_x=\frac{i}{2k}E_{zz}.}
$$
With the <Fourier transform> convention used below and its inverse, this becomes an <ordinary differential equation> for each transverse <wavenumber>:
$$
\partial_x\widehat E(x,\nu)=-\frac{i\nu^2}{2k}\widehat E(x,\nu),\qquad
\boxed{\widehat E(x,\nu)=e^{-i\nu^2x/(2k)}\widehat E(0,\nu).}
$$
Consequently the initial-value solution is
$$
E(x,z)=\int_{\mathbb R}\widehat E(0,\nu)e^{i\nu z-i\nu^2x/(2k)}\,d\nu.
$$
Equivalently, evaluating the oscillatory <Gaussian integral> gives the <one-dimensional transverse Fresnel propagation> formula
$$
\boxed{E(x,z)=\sqrt{\frac{k}{2\pi ix}}\int_{\mathbb R}
\exp\!\left(\frac{ik(z-z')^2}{2x}\right)E(0,z')\,dz',\qquad x>0.}
$$
The square-root branch has $\sqrt{1/i}=e^{-i\pi/4}$. The formula is an <oscillatory integral> for general data, or the <Fresnel propagator> acting on <square-integrable functions>. It approaches the initial field as $x\downarrow0$. A <plane wave> has transverse <wavenumber> $\nu=k\sin\alpha$; the reduced axial phase $-\nu^2x/(2k)$ agrees with $k(\cos\alpha-1)x$ through order $\alpha^2$. This also checks the sign. The <paraxial approximation> applies to $|\nu|\ll k$.
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