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For the free parabolic wave equation $E_x=i{E}_{zz}/(2k)$, input $E(0,z)=e^{-A{z}^2}$ with $\mathrm{Re}A>0$ propagates asThe 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 $q$.

PutIn free space, the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. Under the Fourier transform convention $\hat{E}(p)=\int E(z)e^{-ipz}dz$, it becomesThe Gaussian integral is legitimate because $\mathrm{Re}A=1/D^2>0$. Invert the transform after multiplying by $e^{-i{p}^{2}x/(2k)}$. A second Gaussian integral, or equivalently one-dimensional transverse Fresnel propagation, givesChoose the square-root branch continuously from $q(0{)}^{-1/2}=1$; for real $x\ge 0$ there is no zero of $q$. This gives the correct incident field at $x=0$, and direct differentiation verifies the free parabolic wave equation.

For clarity, the squared envelope magnitude isThe Gaussian beam with one transverse coordinate remains Gaussian, with one-transverse-coordinate amplitude factor $q^{-1/2}$, not the $q^{-1}$ of a beam with two transverse coordinates. The negative initial quadratic phase produces focusing for $F>0$; diffraction prevents a singularity at $x=F$. These expressions describe the paraxial approximation to free propagation, rather than an exact unrestricted Helmholtz equation beam.

Use the time convention $e^{-i\omega t}$. The free-space Helmholtz equation, with $\psi =e^{ikx}E$, gives the exact reduced equationFor a forward plane wave at angle $\alpha$, $E_x=O(k{\alpha }^{2}E)$ and $E_{zz}=O(k^2{\alpha }^{2}E)$, whereas $E_{xx}=O(k^2{\alpha }^{4}E)$. Thus the paraxial approximation neglects $E_{xx}$ relative to $2ik{E}_x$. The reduced field satisfies the parabolic wave equationWith the Fourier transform convention used below and its inverse, this becomes an ordinary differential equation for each transverse wavenumber:Consequently the initial-value solution isEquivalently, evaluating the oscillatory Gaussian integral gives the one-dimensional transverse Fresnel propagation formulaThe 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 $x\downarrow 0$. A plane wave has transverse wavenumber $\nu =k\sin \alpha$; the reduced axial phase $-{\nu }^{2}x/(2k)$ agrees with $k(\cos \alpha -1)x$ through order ${\alpha }^2$. This also checks the sign. The paraxial approximation applies to $|\nu |\ll k$.