Inner product for elements of Fourier basis

Question

Assuming I have a infinite dimensional continuous Hilbert $\mathcal{H}$ space and an orthonormal basis $\{|\,x\,\rangle\}_{\mathbb{R}}$, I want to transform this basis with the Fourier transform to get the new basis $\{|\,p\,\rangle\}_{\mathbb{R}}$. So $$|\,p\,\rangle = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ipx} |\,x\,\rangle\, dx$$Now I want to compute the inner product of two elements in this basis, which according to my material is orthonormal so $\langle\, p\, |\, q\,\rangle = \delta(p - q)$ should hold. But I can't quite follow the calculations. In particular I don't get the following step in the derivation: $$\langle\, p \,| \,q\, \rangle = \dots = \frac{1}{2\pi} \int_{-\infty}^{\infty}e^{-ix(q - p)} dx = \delta(q - p)$$

Answer 1

This is a result from Fourier analysis of distributions.

The Fourier transform of $\delta(x)$ is $$ \mathcal{F}\{\delta(x)\} = \int_{-\infty}^{\infty} \delta(x) \, e^{-ix\xi} \, dx = 1. $$

By the Fourier inversion theorem, if $\mathcal{F}\{f(x)\} = g(\xi)$ then $\mathcal{F}\{g(x)\} = 2\pi \, f(-\xi)$. This is valid also for distributions, so applying it on the former result, we get $$ \mathcal{F}\{1\} = \int_{-\infty}^{\infty} 1 \, e^{-ix\xi} \, dx = 2\pi\,\delta(-\xi) = 2\pi\,\delta(\xi) . $$

NB. Writing these Fourier transforms of distributions as integrals is actually abuse of notation. But it's an abuse that is common in physics.