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I was reading on squeezed Gaussian states and stumbled upon this paper:

Equivalence Classes of Minimum-Uncertainty Packets. II.

It is mentioned after Eq. $\left(2\right)$ that $$ \left\langle x\left\vert\,{{\rm e}^{{\rm i}rxp}}\,\right\vert y\right \rangle = \delta\left(x{\rm e}^{r} - y\right) $$ where $\left\vert x \right\rangle , \left\vert y \right\rangle$ are position eigenstates.

How is this relation derived?

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    $\begingroup$ It would make it easier for others to answer if you gave more background behind your question, especially since the paper you linked is behind a paywall. $\endgroup$ – J.V.Gaiter May 31 at 18:55
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OP's equation $$\begin{align} \langle x | e^{ir\hat{x}\hat{p}/\hbar}| y\rangle ~=~~& \int_{\mathbb{R}} \mathrm{d}p ~\langle x | e^{ir\hat{x}\hat{p}/\hbar}| p\rangle ~\langle p | y\rangle \cr ~\stackrel{(B)+(D)}{=}&~ \int_{\mathbb{R}} \mathrm{d}p ~\frac{e^{ixe^rp/\hbar}}{\sqrt{2\pi\hbar}}~\frac{e^{-iyp/\hbar}}{\sqrt{2\pi\hbar}}\cr ~=~~& \delta(xe^r\!-\!y) \end{align}\tag{A}$$ follows from $$ \langle x | e^{ir\hat{x}\hat{p}/\hbar}| p\rangle~=~\frac{e^{ixe^rp/\hbar}}{\sqrt{2\pi\hbar}}.\tag{B}$$ Here we have assumed the CCR $$ [\hat{x},\hat{p}]~=~i\hbar \hat{\bf 1},\tag{C}$$ and the standard Fourier-transform overlap $$ \langle x | p\rangle ~=~ \frac{e^{ixp/\hbar}}{\sqrt{2\pi\hbar}},\tag{D}$$ cf. e.g. this Phys.SE post.

Sketched proof of eq. (B): Firstly, note that (B) is true for $r=0$. Secondly, differentiate both sides of eq. (B) wrt. $r$ and see that they satisfy the same first-order ODE. Use that $$ e^{-ir\hat{x}\hat{p}/\hbar}\hat{p} e^{ir\hat{x}\hat{p}/\hbar}~=~ e^{-ir[\hat{x}\hat{p}/\hbar, \cdot]}\hat{p}~\stackrel{(C)}{=}~e^r\hat{p}.\tag{E}$$ $\Box$

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