Wave propagation in free space in the limit $d\to 0$

Question

In Goodman's Introduction to Fourier Analysis, the evolution of the field amplitude $U(x_0)$ of a wave propagating in free space is described by the Fresnel approximated operator $$\mathcal R[d]\{U(x_0)\} =\frac{1}{\sqrt{i\lambda d}} \int dx_0 \ U(x_0) e^{\frac{ik}{2d}(x_1-x_0)^2}, $$ where $k,\lambda$ are the wavenumber and the wavelength and $d$ is the travelled distance. Let me rewrite this expression by introducing the propagator $T(x, x', d):=(i\lambda d)^{-\frac{1}{2}}\exp\left(\frac{ik}{2d}(x-x')^2\right)$ so that the evolved amplitude is $$U'(x_1, d)=\int dx_0 \ T(x_0,x_1,d)U(x_0). $$

I would expect that $\lim_{d\to 0}U'(x_1, d)=U(x_1)$ as there is no actual propagation through space. But this should mean that $\lim_{d\to 0}T(x_0, x_1,d)=\delta(x_0-x_1)$, which is not immediately obvious from the definition of $T$. Can anyone help me to prove this?

Answer 1

It is. In fact it happens that $$\delta(x)=\lim_{n\rightarrow \infty}\sqrt{\frac{n}{\pi}}e^{-nx^2}$$ You can convince yourself by noting that the exponential goes to zero much faster than the square root to infinity, except if the exponent is 0. Then you can substitute $d\sim \frac{1}{n}$ and get your Delta function.