 Solution

Free propagation is convolution with a chirp, $K_L(x)=C_L{e}^{ikL}{e}^{ik{x}^2/(2L)}$. The convolution theorem therefore turns it into multiplication by its Fourier transform. Complete the square:The Gaussian integral then givesThus the wavevector-dependent phase isup to an overall phase convention; after removing the carrier it is $-L{k}_{\perp }^2/(2k)$. This also follows from $k_z=\sqrt{k^2-k_{\perp }^2}\simeq k-k_{\perp }^2/(2k)$.

Normalization qualification: the source keeps a two-dimensional-style prefactor while integrating over only one transverse coordinate. With its literal $C_L=ik/(2\pi L)$, there is an additional $q$-independent factor, rather than solely a unit-modulus multiplier. A properly normalized one-dimensional Fresnel propagator has $C_L=\sqrt{k/(2\pi iL)}$, which cancels the square-root factor above and makes propagation a pure Fourier-domain phase. For example, the literal source kernel multiplies a plane-wave Fourier mode by a factor of magnitude $\sqrt{k/(2\pi L)}$, not one. The spatially dependent phase and the proportional Fourier-transform results remain unchanged. ^[ref]