On the physical meaning of the probabilistic approach of quantum mechanics

Question

In Griffith's introduction to quantum mechanics, one of the problems asks to find the expectation value of the position of a particle at state

$$\psi(x,t)=A\mathrm e^{(-a((mx^2)/\hbar+\mathrm it))}.$$

After working the integral, the result is that $\langle x\rangle =0$ because it comes down to integrating an odd function over a symmetric interval which yields zero. but what does that mean? what does it mean that the expectation value of the particle's position is zero when we're integrating over all of space? Doesn't the particle need to be somewhere in space and hence have some finite nonzero expectation value?

The same thing also happens when trying to find the expectation value for the momentum of the particle at the same state.

Answer 1

Probabilistic approach means that sometimes we will find the particle in one place, $x_1$, sometimes in another, $x_2$, but on average its position will be zero: $$ \bar{x} = \frac{1}{N}\sum_{i=1}^N $$ More precisely, this some tends to zero as $N$ goes to infinity. The limit of $N\rightarrow +\infty$ is then given by $\langle x\rangle$.

This is not the end of the story however. The measurement postulate of quantum mechanics, which tells us that by measuring position we change the particle wave function, localizing the particle in a particular place. This means that, we cannot make $N$ measurements on the same particle, but rather we do measurements on $N$ identical particles "prepared" in the same way, i.e., having the same initial wave function. Such ensemble averaging is always implicit in quantum mechanics, although it often seems, as if we are talking about only one particle. (Incidentally, this replacing of multiple measurements on a single particle by ensemble measurements is behind some of the debates about the nature of quantum mechanics.)