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

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?

• "which is not immediately obvious" "Can anyone help me", we do want to help you, but what kind of help are you looking for? It is typically not obvious when a certain limit ends up being the Dirac delta distribution; you simply realise that the finite case satisfies sufficient properties that you understand that the limit must end up being Dirac delta distribution, and be done with that. We just collect all the different ways for functions to limit towards the Dirac delta distribution into a gigantic table. Commented Jun 2, 2023 at 11:39
• @naturallyInconsistent I thought someone here might be able to provide a mathematical proof, that's all. Commented Jun 2, 2023 at 12:02
• So, then, what would constitute a mathematical proof of that to you? Commented Jun 2, 2023 at 12:19
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
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Commented Jun 2, 2023 at 13:02
• @naturallyInconsistent I cannot get any clearer than this. I'd like to prove that $\lim_{d\to 0}T(x_0, x_1,d)=\delta(x_0-x_1)$ as defined above. There is already a helpful answer below. Commented Jun 2, 2023 at 13:17

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.