Fresnel integral: stationary phase approx.

Question

The following integral describes the propagation of light (in certain cases) $$ U_i(x,y,z) = \frac{e^{ikz}}{i\lambda z} \int_{-\infty}^\infty d\xi \int_{-\infty}^\infty d\eta \; U_0(\xi, \eta,0) \, \exp\left({i \frac{k}{2z}\left[ (x-\xi)^2 + (y-\eta)^2 \right] }\right) $$ In the limit $z\to 0$ it is said that the quadratic phase factor act like a Dirac $\delta$ distribution. While this is intuitively clear (in the context of optics) I don't see how it works mathematically.

• $k=2\pi/\lambda > 0$ is the wave number,
• $\lambda>0$ is the wavelength,
• $(x,y,z)$ is a position, and so is $(\xi, \eta, 0)$. Hence, they are all real numbers.
• $U_0$ is a real valued function, which represent the electric field. It is a "well-behaved" function (i.e. finite, and smooth). Usually we consider that it differs from zero only on a finite surface $\Sigma$ -- the aperture.

Answer 1

1. We can write $U_i$ as a convolution between two function, $U_0$ and $h(x,y,z) = \ldots$ $$ U_i(x,y,z) = U_0(x,y,0) * h(x,y,z) $$
2. Next, we Fourier transform both sides (in 2D) and use the Fourier theorem for convolutions. This yields $$ \tilde U_i(f_x, f_y, z) = \tilde U_0(f_x,f_y, 0) \cdot \tilde h(f_x,f_y,z) $$ Doing the calculation, we get $ \tilde h(f_x,f_y,z) = e^{ikz} e^{-i\pi \lambda z (f_x^2 + f_y^2)} $.
3. Now, we take the limit $z\to 0$ and obtain $\tilde h(f_x,f_y,0) = 1$.
4. Finally, we use the inverse 2D Fourier transform on both sides as well as the Fourier theorem for convolutions. This yields $$ U_i(x,y,z) = U_0(x,y) * \mathcal{F}^{-1}\{1\} = U_0(x,y) * \delta(x,y) $$