---
tags: Notes on Noise
---
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Début préambule à ne pas modifier
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$$
\def\moy#1{{\langle #1 \rangle}}
\def\abs#1{{\left|#1\right|}}
\def\dd{\mathrm{d}}
\def\ed{\mathrm{e}}
\def\norm#1{{\parallel #1 \parallel}}
\def\ket#1{{| #1 \rangle}}
$$
# Remarques sur l'Eq. (B2)
To simplify the discussion, I rewrite the Eq. (B2) in a smpler form as
$$
I(q) = \int f(k)e^{ikq} \exp(-k^2\Delta^2) dk.
$$
I think that the conclusion written after Eq. (B2) in the paper :
*"For $\Delta^{-1}$ larger than the width of f(k) (which means that the width of $P(q)$ should be much smaller than the the one of $\tilde{f}(q)$, the noise can be deconvoluted (of course, if one assumes its Gaussian form)"*
it is not enough to ensure that I(q) is equal to $\int f(k)e^{ikq} dk$

In my opinion, $\Delta^{-1}$ must be larger than the highest value of $k$ in the support of $f(k)$
Let's see an example:
suppose that $f(k)$ is not zero in the interval $[k_0,k_c]$. Then the condition for the integral $I(q)$ to be not modified by the gaussian kernel $\exp(-k^2\Delta^2)$ is $\Delta^{-1} \gg k_c$, which is different than $\Delta^{-1} \gg k_c-k_0$, as I understand the meaning of *"$\Delta$ larger than the width of f(k) "* written in the paper.

Therefore we end with the conclusion that $\Delta$ must be much smaller than 1/k_c. (or $1/\omega_c$ if $q$ is the time).