We know that $x_0$ and $p_0$ are the mean values for the position and momentum of a particle in the normalized state characterized by the function $\psi (x)$, ( that is, $x_0=\langle x \rangle_\psi$ and $p_0=\langle p \rangle_\psi$).
Does the mean value of $x$ change for the function $\psi(x+x_0)$ ?
If the mean value for x for the funtion $\psi(x)$ is:
$$\langle x \rangle_\psi=\int_{-\infty}^\infty \psi^\ast (x)\; x\; \psi(x) dx=x_0$$
The mean value for x for the function $\psi(x+x_0)$ is:
$$\langle x \rangle_{\psi' }=\int_{-\infty}^\infty \psi^\ast(x+x_0)\; x\; \psi(x+x_0) dx$$
If we center the function in $x=x_0$, we have:
$$\langle x \rangle_{\psi' }=\int_{-\infty}^\infty \psi^\ast(x)\; (x-x_0)\; \psi(x) dx$$
Then
$$\langle x \rangle_{\psi' }=\int_{-\infty}^\infty \psi^\ast(x)\; x\;\psi(x)dx -\int_{-\infty}^\infty \psi^\ast(x)\;x_0\; \psi(x) dx$$
Where the first integral is $\langle x \rangle_\psi$ and the second one, we know that
$$\int_{-\infty}^\infty \psi^\ast(x)\psi(x) dx=1$$
Because it is normalized, so:
$$\langle x \rangle_{\psi' }= \langle x \rangle_\psi -x_0$$
Where $\langle x \rangle_\psi=x_0$, then:
$$\langle x \rangle_{\psi' }=0$$
Is that correct?
If so, what will be the mean value for $p$ with $\psi(x+x_0)$?
