# Scaled Dirac Delta function: $\delta (xe^r - y)$

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

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?

• 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. – J.V.Gaiter May 31 at 18:55

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 $$e^{-ir\hat{x}\hat{p}/\hbar}\hat{p} e^{ir\hat{x}\hat{p}/\hbar}~=~ e^{-ir[\hat{x}\hat{p}/\hbar, \cdot]}\hat{p}~\stackrel{(C)}{=}~e^r\hat{p}.\tag{E}$$ $$\Box$$