If I look at a gaussian wave package, and then interpret (in the usual Quantum-Mechanics Way) its square value as the propability density, then I can calculate a mean value for the position: $$ x_{mean}(t) = \int \Psi^*(x,t) x \Psi(x,t) dx $$
Is the time derivative of this mean position the same as the group velocity of the wave package, that means, is $ \dot{x}_{mean} = \frac{d \omega}{ d k}$?
I tried to derive it (because this is what I imagine the group velocity to be), but I failed.
Group velocity is a property of a linear superposition of plane wave solutions of the form $e^{iq\cdot x-i\omega t}$, with dispersion relation $\omega(q)$ that relates frequency to wave number. From this, you can assume $\psi(x,t)=\int\frac{d^3q}{(2\pi)^3}g(q)e^{iq\cdot x-i\omega(q)t}$ for some initial (normalized) wave function $g(q)$. For your question, you can assume that $g(q)$ is a Gaussian sharply peaked near some frequency $q_0$.
Also, $xe^{iq\cdot x}=-i\frac{d}{dq}e^{iq\cdot x}$.
ANSWER \begin{align*} \langle x \rangle(t) &= \int d^3x' \bigg[x'\int \frac{d^3qd^3q'}{(2\pi)^6}g(q)g^*(q')e^{i(q-q')\cdot x'-i(\omega(q)-\omega(q'))t} \bigg]\\ &=\int d^3x'\bigg[\int\frac{d^3 qd^3 q'}{(2\pi)^6}g(q)g^*(q')(-i\nabla_q e^{i(q-q')\cdot x'})e^{-i(\omega(q)-\omega(q'))t}\bigg]\\ &=\int d^3x'\bigg[\int\frac{d^3qd^3q'}{(2\pi)^6}\Big(i\nabla_qg(q)g^*(q')+\nabla_q\omega(q)tg(q)g^*(q')\Big)e^{i(q-q')\cdot x-i(\omega(q)-\omega(q'))t}\bigg]\\ &\text{(from integration by parts)}\\ &= \int\frac{d^3qd^3q'}{(2\pi)^6}(2\pi)^3\delta^3(q-q')\Big(i\nabla_q g(q)g^*(q')+\nabla_q\omega(q)tg(q)g^*(-q)\Big)e^{-i(\omega(q)-\omega(q'))t}\\ &\text{(using the momentum-space delta function with Phys. convention)}\\ &=\int \frac{d^3 q}{(2\pi)^3}\Big(i\nabla_q g(q)g^*(q)+\nabla_q\omega(q)|g(q)|^2t\Big)\\ &\approx\nabla_q\omega(q_0)t + \mathcal O(1=t^0) \\ &\text{(assuming $g(q)$ is sharply peaked about $q_0$)} \end{align*} Hence, $\lim_{t\rightarrow\infty}\frac{\langle x\rangle(t)}{t}=\nabla_q\omega(q_0)$
