Fourier Transforms

The Dirac delta function has the following integral representation

(1)
\begin{align} \int \frac{dk}{2\pi} e^{-ik(x-x')} = \delta(x-x')\ . \end{align}

This can be interpreted as the completeness relation in the $x$-basis

(2)
\begin{align} \int dk |k\rangle \langle k| = \mathbf{1} \implies \int dk \langle x|k\rangle \langle k|x'\rangle = \delta(x-x')\ , \end{align}

where $\langle x| k\rangle = e^{-ikx}/\sqrt{2\pi}$.

(3)
\begin{align} \psi(x)\equiv \langle x | \psi\rangle = \int dk \langle x | k \rangle \langle k|\psi \rangle = \int \frac{dk}{\sqrt{2\pi}}\ e^{-ikx} \widetilde{\psi}(k)\ , \end{align}

where $\widetilde{\psi}(k)\equiv \langle k|\psi \rangle$. It is thus easy to see that two representations of the state $|\psi\rangle$ in the x-basis $\psi(x)$ and in the k-basis $\widetilde{\psi}(k)$ are indeed Fourier transforms of each other.