How to calculate the expectation value of position vector? $$\psi (\vec{x})=Ae^{-(1/4a^2)|\vec{x}-\vec{x}_0|^2}e^{i\vec{p}_0\cdot \vec{x}/\hbar}$$
The wave function is like this, then how is the expectation value of position vector (not position) calculated?
 A: The mean value of the position is given by $\langle \psi|\mathbf{X}|\psi\rangle$. Now insert the completeness relation for position states to get
$$\langle \psi|\mathbf{X}|\psi\rangle=\int_{\mathbb{R}^3} \mathbf{x} |\psi(\mathbf{x})|^2\,\mathrm{d}V$$
From the OP, we see that the modulus squared of the wave function is
$$|\psi(\mathbf{x})|^2=A^2\exp\left(-\frac{|\mathbf{x}-\mathbf{x}_0|^2}{2a^2}\right)$$
Now we insert this into the above integral:
$$\langle \psi|\mathbf{X}|\psi\rangle=A^2\int_{\mathbb{R}^3} \mathbf{x}\exp\left(-\frac{|\mathbf{x}-\mathbf{x}_0|^2}{2a^2}\right) \,\mathrm{d}V$$
Define $\mathbf{y}=\mathbf{x}-\mathbf{x}_0$,
$$\langle \psi|\mathbf{X}|\psi\rangle=A^2\int_{\mathbb{R}^3}(\mathbf{y}+\mathbf{x}_0)\exp\left(-\frac{|\mathbf{y}|^2}{2a^2}\right) \,\mathrm{d}V'$$
This splits into two terms. The first term is odd and vanishes when integrated. The second is proportional to $\mathbf{x}_0$. The constant factor will cancel the $A^2$ when properly normalized. (It is $(a\sqrt{2\pi})^3$.) Thus
$$\langle \psi|\mathbf{X}|\psi\rangle=\mathbf{x}_0$$
