Gaussian envelope operator applied to an arbitrary wave function I am studying the reference given in  [1]. In that, the authors define the non-unitary "Gaussian envelope" quantum operator $\mathcal{\hat{M}}$ (see Eq. 4 in reference [1]):
$$\mathcal{\hat{M}} = C \int_{-\infty}^{+ \infty} dq~e^{-q^2 \Omega^2/2}\left|q \right> \left<q \right|, \tag{1}$$
where $\left|q \right>$ is an eigenstate of the position basis; that is $\hat{q}' \left|q \right>=q\left|q \right>$ and $C$ is a constant of proportionality. Besides, I think that is important to mention that the $\left\lbrace \left|q \right>\right\rbrace$ basis is related to the conjugate (momentum) basis $\left\lbrace \left|p \right>\right\rbrace$ by a Fourier transform; then, using the convention of the good reference [2] (see Eqs. (9)), we mean
$$\left|q \right>=\left(2\sqrt{\pi} \right)^{-1}\int_{-\infty}^{+\infty} dp~e^{-iqp/2} \left|p \right>, \tag{2}$$
$$\left|p \right>=\left(2\sqrt{\pi} \right)^{-1}\int_{-\infty}^{+\infty} dp~e^{+iqp/2} \left|q \right>. \tag{3}$$
If we apply the operator given in Eq. (1) to an arbitrary state $\left|\psi \right>$, we have
$$\mathcal{\hat{M}}\left|\psi \right> = C \int_{-\infty}^{+ \infty} dq~e^{-q^2 \Omega^2/2}\left|q \right> \left<q \right|\left. \psi\right> $$
$$=C \int_{-\infty}^{+ \infty} dq~\psi_{G}(q)~ \psi(q)\left|q \right>, \tag{4}$$
where $\left<q \right|\left. \psi\right>=\psi(q)$ and we condensate the Gaussian wave function function $\psi_{G}(q)=e^{-q^2 \Omega^2/2}$, which have a variance of  $1/2\Omega^2$ as can be directly verified from the squared modulus $\left|\psi_{G}(q) \right|^2$. Then, according to reference  [1], the last line of Eq. (4) is equivalent to convolution in momentum space by a Gaussian with variance $\Omega^2/2$ (see the argument below of Eq. (4) in reference  [1]). So, my question is:
Can someone give me a hint to mathematically prove this argument?
References:

*

*N. C. Menicucci et al., Universal Quantum Computation with Continuous-Variable Cluster States, Phys. Rev. Lett. 97, 110501 (2006), arXiv:quant-ph/0605198.

*C. Weedbrook et al., Gaussian quantum information, Rev. Mod. Phys. 84, 621 (2012), arXiv:1110.3234.

 A: With the hints showed in the comments, I am able to answer my own question. First we stablish the relation between the components of the arbitrary state $\left| \psi \right>$ in a the position or momentum basis:
$$\left<p\right|\left. \psi \right>=\psi(p)=(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dq~\psi(q)e^{-iqp/2}=\mathcal{F}[\psi(q)], \tag{1}$$
$$\left<q\right|\left. \psi \right>=\psi(q)=(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~\psi(p)e^{iqp/2}=\mathcal{F}^{-1}[\psi(p)], \tag{2}$$
which can be proved inserting the identity $\mathbb{\hat{I}}=\int_{-\infty}^{+\infty} dq~\left| q\right> \left<q \right|=\int_{-\infty}^{+\infty} dp~\left| p\right> \left<p \right|$ in $\left<p\right|\left. \psi \right>$ and $\left<q\right|\left. \psi \right>$ and using the definitions of Eqs. (2) and (3) of the question. On the other hand, the Eqs. (1) and (2) above express correctly the definitions for the Fourier transform between the components of $\left|\psi\right>$ in position or momentum space.
Now, we denote the product of functions of the last line in Eq. (4) in the question as $\left< q\right| \left.\Psi \right>=\psi_{G}(q) \psi(q)$; therefore, such state can be written as
$$\mathcal{\hat{M}}\left|\psi \right>=C\int_{-\infty}^{+\infty}dq~ \left< q\right| \left.\Psi \right> \left|q \right>; \tag{3}$$
then, insert the identity and develop:
$$\mathcal{\hat{M}}\left|\psi \right>=C\int_{-\infty}^{+\infty}dq~ \left< q\right|\mathbb{\hat{I}} \left|\Psi \right> \left|q \right>$$
$$=C\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}dq dp~ \left< q\right|\left. p\right>\left<p \right.\left|\Psi \right> \left|q \right>$$
$$=C (2\sqrt{\pi})^{-1}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}dq dp~ e^{+iqp/2}~\Psi(p)\left|q \right>$$
$$=C \int_{-\infty}^{+\infty}dp~\Psi(p) \left[ (2\sqrt{\pi})^{-1}\int_{-\infty}^{+\infty} dq ~e^{+iqp/2}\left|q \right>\right]$$
$$=C \int_{-\infty}^{+\infty}dp~\Psi(p) \left|p \right>, \tag{4}$$
where we have used  $\left< q\right|\left. p\right>=\left( 2\sqrt{\pi}\right)^{-1}e^{+iqp/2}$ (see Eqs. (2) and (3) from the question) and $\left< p\right|\left. \Psi \right>= \Psi(p)$. Notably, the components $\Psi(p)$ can be obtained from the ones in the position representation as Eq. (1) establishes; therefore,
$$\Psi(p)=\mathcal{F}[\Psi(q)]=\mathcal{F}[\psi_{G}(q) \psi(q)]. \tag{5}$$
Now we consider the convolution theorem:
$$\mathcal{F}\left[f \cdot  g\right]=\mathcal{F}[f] \ast \mathcal{F}[g], \tag{6}$$
where $(\cdot)$ denotes point-wise multiplication and $(\ast)$ represents convolution. Then, applying this theorem in Eq. (4) we have
$$\mathcal{\hat{M}}\left|\psi \right>=C \int_{-\infty}^{+\infty}dp~\mathcal{F}\left[\psi_{G}(q) \psi(q) \right]\left|p \right>,$$
$$=C \int_{-\infty}^{+\infty}dp~\mathcal{F}\left[\psi_{G}(q)\right] \ast \mathcal{F}\left[ \psi(q) \right]\left|p \right>; \tag{7}$$
besides, we have
$$\mathcal{F}\left[\psi_{G}(q)\right]=e^{-p^2/2\Omega^2},$$
$$\mathcal{F}\left[ \psi(q) \right]=\psi(p),$$
therefore, the state of Eq. (7) is
$$\mathcal{\hat{M}}\left|\psi\right>=C' \int_{-\infty}^{+\infty}dp~\left[e^{-p^2/2\Omega^2}\ast \psi(p)\right]\left|p \right>; \tag{8}$$
which constitutes the desired result.
