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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:

  1. N. C. Menicucci et al., Universal Quantum Computation with Continuous-Variable Cluster States, Phys. Rev. Lett. 97, 110501 (2006), arXiv:quant-ph/0605198.
  2. C. Weedbrook et al., Gaussian quantum information, Rev. Mod. Phys. 84, 621 (2012), arXiv:1110.3234.
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    $\begingroup$ Try inserting the identity in momentum space $\endgroup$
    – FizzKicks
    Nov 23 '21 at 3:10
  • $\begingroup$ This question is much simpler than it looks. You're just dealing with the fact $\mathcal{F}(f * g) = \mathcal{F}(f) \otimes \mathcal{F}(g)$ where here $\mathcal{F}$ means Fourier transform and $\otimes$ means convolution. $\endgroup$
    – DanielSank
    Nov 23 '21 at 20:14
  • $\begingroup$ @DanielSank, I get it!. The Fourier transform of a point-wise multiplication of the functions $f$ and $g$ is the convolution between the individual Fourier transforms. $\endgroup$ Nov 24 '21 at 2:18
  • $\begingroup$ @JulioAbrahamMendozaFierro correct. This is not a "quantum" thing at all, but rather a very useful fact of mathematics. By the way, if you can think of a Fourier transform as a change of basis, you will enjoy peace in your life. Think of the values of a function $f(x)$ as the components of a vector $\lvert f \rangle$ in a certain basis, i.e. $\langle x | f \rangle = f(x)$, and think of the values of that function's Fourier transform $\tilde{f}(k)$ as the components of $\lvert f \rangle$ in a different basis, i.e. $\langle k | f \rangle = \tilde{f}(k)$. $\endgroup$
    – DanielSank
    Nov 24 '21 at 19:14
  • $\begingroup$ Remember that $\int dx \lvert x \rangle \langle x \rvert$ is the identity and use that fact (and the equivalent one with $k$) to understand the Fourier transform (noting that $\langle x | k \rangle = \exp(i k x)$). $\endgroup$
    – DanielSank
    Nov 24 '21 at 19:15
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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.

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