Double differentiation of characteristic function of Normal random variable Knowing that a one-dimensional random variable $\Gamma$ is Gaussian if it has the characteristic function
$$\mathbb{E}\hspace{0.15cm}e^{i\xi\Gamma}=e^{im\xi-\frac{1}{2}\sigma^2\xi^2}\tag{1}$$
for some real numbers $m\in\mathbb{R}$ and $\sigma\geq0$. If we differentiate $(1)$ two times with respect to $\xi$ and set $\xi=0$, we see that
$$m=\mathbb{E}\hspace{0.15cm}\Gamma\hspace{0.3cm}\sigma^2=\mathbb{V}\hspace{0.15cm}\Gamma\tag{2}$$

I cannot understand how to get to $(2)$ by double differentiating $(1)$ with respect to $\xi$, setting $\xi=0$. If I differentiate r.h.s. of $(1)$ two times with respect to $\xi$ and I set $\xi=0$, I get $-m^2$.
This question, the other answer, and the comment discussion are muddled.
For a random variable $X$ with finite second moment, with characteristic function  $\varphi_X(t)=E[e^{itX}]$, the following are true:
$$  \varphi_X(t)=1+ iE[X]t -\frac{E[X^2]}2 t^2+o(t^2)\\E[X] =\mu = -i\varphi_X'(0)\\E[X^2] = \mu^2+\sigma^2= -\varphi_X''0).$$
If you define $\psi=\log\varphi_X$, then
$$\psi(t)=\mu it-\sigma^2\frac{t^2}2+o(t^2)\\
\mu = -i \psi'(0)\\
\sigma^2 = -\psi''(0),$$
which can be connected with the previous equations by application of the chain rule and product rule:
$$\varphi_X(t)=e^{\psi(t)}\\
\varphi_X'(t) = e^{\psi(t)} \psi'(t)\\
\varphi_X''(t) = e^{\psi(t)}( \psi'(t))^2 + e^{\psi(t)}\psi''(t)$$$$
\varphi_X'(0) = e^{\psi(0)} \psi'(0)=\psi'(0)=-i\mu\\
\varphi_X''(0) = e^{\psi(0)}( \psi'(0))^2 + e^{\psi(0)}\psi''(0)=-\mu^2-\sigma^2,$$
and so on.
I will treat $\mathbb{E}$ as a constant.
$$\mathbb{E} \cdot e^{iξ \Gamma}=e^{imξ−\frac{1}{2}σ^2ξ^2}$$
So then
$$\mathbb{E} \cdot e^{iξ \Gamma}=\frac{e^{imξ}}{e^{\frac{1}{2}σ^2ξ^2}}$$
Differentiating once w.r.t $ξ$,
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \cdot im \cdot  e^{imξ} - e^{imξ} \cdot 2 \cdot\frac{1}{2}\sigma^2 \cdot ξ \cdot e^{\frac{1}{2}σ^2ξ^2}}{\bigg( e^{\frac{1}{2}σ^2ξ^2} \bigg)^2}$$
Simplifying, we obtain:
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \cdot im \cdot  e^{imξ} - e^{imξ} \cdot \sigma^2 ξ \cdot e^{\frac{1}{2}σ^2ξ^2}}{e^{σ^2ξ^2}}$$
Furthermore,
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \bigg( im \cdot  e^{imξ} - e^{imξ} \cdot \sigma^2 ξ \bigg) }{\bigg (e^{\frac{1}{2} \cdot σ^2ξ^2} \bigg)^2}$$
So
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{im \cdot  e^{imξ} - e^{imξ} \cdot \sigma^2 ξ}{e^{\frac{1}{2} \cdot σ^2ξ^2}}$$
Which means that
$$\mathbb{E} i\Gamma \cdot e^{iξ \Gamma} \cdot e^{\frac{1}{2} \cdot σ^2ξ^2} =e^{imξ} (im - \sigma^2 ξ)$$
Multiplying both sides by $e^{-im ξ}$,
$$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)} =im - \sigma^2 ξ$$
Differentiating again w.r.t  $ξ$ we obtain:
$$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)} \cdot \frac{d}{dξ}\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)= - \sigma^2$$
Thus,
$$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)} \cdot\bigg( i \Gamma + 2\frac{1}{2} \cdot σ^2ξ - im\bigg)= - \sigma^2$$
Finally, we obtain that if we differentiate the first equation twice by $ξ$ we obtain that:
$$\mathbb{E} i\Gamma \cdot e^{ iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ} \cdot\bigg( i \Gamma + σ^2ξ - im\bigg)= - \sigma^2$$
Now setting $$ξ=0,$$
We obtain that:
$$\mathbb{E} i\Gamma \cdot e^{ i(0) \Gamma + \frac{1}{2} \cdot σ^2(0)^2 - im  (0)} \cdot\bigg( i \Gamma + σ^2(0) - im\bigg)= - \sigma^2$$
So then
$$\mathbb{E} i\Gamma \cdot e^{0} \cdot\bigg( i \Gamma- im\bigg)= - \sigma^2$$
Thus,
$$\mathbb{E} i\Gamma \cdot\bigg( i \Gamma - im\bigg)= - \sigma^2$$
Simplifying our expression,
$$\mathbb{E} i^2\Gamma^2 - \mathbb{E}i^2 \Gamma m= - \sigma^2$$
$$-\mathbb{E}\Gamma^2 + \mathbb{E}\Gamma m= - \sigma^2$$
Then
$$\mathbb{E}\Gamma m= \mathbb{E}\Gamma^2- \sigma^2$$
$$ m= \frac{\mathbb{E}\Gamma^2- \sigma^2}{\mathbb{E}\Gamma}$$
So
$$ m= \Gamma - \frac{\sigma^2}{\mathbb{E}\Gamma}$$
You should note that I treat $\mathbb{E}$ as a constant. If it is a function of $ξ$, I will edit my answer.
But since RHS isn't dependent on $\mathbb{E}$, I can safely say that the second derivative of the RHS is
\begin{align*}
\frac{d^2}{dξ^2} \bigg( e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) &amp;=  
\frac{d}{dξ}\bigg[ \bigg( e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) \cdot \bigg( im - σ^2ξ  \bigg) \bigg] \\ 
&amp;= \frac{d}{dξ}\bigg(  im \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - σ^2ξ \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) \\
&amp;= im \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \bigg[ \sigma^2ξ \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} + \sigma^2 \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg] \\
&amp;= im \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \sigma^2ξ \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \sigma^2 \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \\
&amp;= e^{imξ−\frac{1}{2}σ^2ξ^2} \cdot \bigg[ im \big( im - \sigma^2 ξ \big) -  \sigma^2ξ \big( im - \sigma^2 ξ \big) - \sigma^2 \bigg]
\end{align*}
Now setting $ ξ=0$ we obtain that:
$$\frac{d^2}{dξ^2} \big( e^{imξ−\frac{1}{2}σ^2ξ^2} \big) \bigg|_{ξ=0} =
i^2m^2 - \sigma^2 = -m^2 - \sigma^2$$