$\newcommand{\ket}[1]{| #1\rangle}$This question basically has two very related parts. This came up in the context of trying to verify something my professor said a while ago: that if the wave functions for two identical particles are well separated (i.e. if they are very peaked, and the peaks are macroscopically far apart), then you can model them as distinguishable particles. He reasoned that when we swap the wave functions, the term we get from swapping is very small and can be ignored. So, taking the norm of the symmetrized wave function reduces to taking the norm of the non-trivial amplitude in the symmetrization. This will be the amplitude you would get by modeling the particles as distinguishable.
Yet, when I do this procedure, the normalization from the symmetrization procedure screws me up. I am unsure where I am going wrong.
Suppose I have two bosons. I know that one is in the state $\ket{\psi_1}$ and one is in the state $\ket{\psi_2}$.
The symmetrized state then is
$$ \frac{1}{\sqrt{2}} \left(\ket{\psi_1 \psi_2} + \ket{\psi_2\psi_1} \right)$$
Suppose the wave functions for $\ket{\psi_1}$ and $\ket{\psi_2}$ are peaked at separated place, or have non-overlapping support.
Then, if I look at $x_1,x_2$, with $x_1 \in \text{supp}({\psi_1 (\cdot)})$ and $x_2 \in \text{supp}({\psi_2 (\cdot)})$, then won't the probability that I observe a particle near $x_1$ and another near $x_2$ be
$$ \frac{1}{2} |\psi_1(x_1)|^2 |\psi_2(x_2)|^2$$
which is half of what it would be for distinguishable particles? This result feels wrong. I thought maybe something was wrong with the normalization, but it doesn't seem to be that there is.