= Solution
Free propagation is <convolution> with a chirp, $K_L(x)=C_Le^{ikL}e^{ikx^2/(2L)}$. The <convolution theorem> therefore turns it into multiplication by its <Fourier transform>. Complete the square:
$$
\frac{kx^2}{2L}-qx=\frac{k}{2L}\left(x-\frac{Lq}{k}\right)^2-\frac{Lq^2}{2k}.
$$
The <Gaussian integral> then gives
$$
\widetilde E(q,L)=C_L\sqrt{\frac{2\pi iL}{k}}\,
e^{ikL-iLq^2/(2k)}\widetilde E(q,0).
$$
Thus the wavevector-dependent <phase> is
$$
\boxed{\varphi(k_\perp)=kL-\frac{Lk_\perp^2}{2k},}
$$
up to an overall <phase> convention; after removing the carrier it is $-Lk_\perp^2/(2k)$. This also follows from $k_z=\sqrt{k^2-k_\perp^2}\simeq k-k_\perp^2/(2k)$.
\b[<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. <Oscillations and Waves by Richard Fitzpatrick>[Wave optics, two-dimensional Fourier optics: optical-path expansion and Fourier-domain diffraction]{ref}
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