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Let's say we have two identical particles, $r_1$ is the position of the first particle and $r_2$ is the position of the second particle. The wave function is $\psi(r_1,r_2)$. Since these particles are identical it has to be true that $|\psi(r_1,r_2)|^2=|\psi(r_2,r_1)|^2$. Thus $|\psi(r_1,r_2)|^2dr_1dr_2$ is the probability of finding particle $1$ in the volume $dr_1$ and particle $2$ in the volume $dr_2$. Is it correct to tell that the probability that one particle is the volume $dr_1$ and the other in the volume $dr_2$ is: $|\psi(r_1,r_2)|^2dr_1dr_2+|\psi(r_2,r_2)|^2dr_1dr_2=2|\psi(r_1,r_2)|^2dr_1dr_2$

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I believe the elementary probability to find the two particles (indistinguishably) around $r_1$ and $r_2$ is indeed $2 \left| \psi(r_1, r_2) \right|^2 dr_1 dr_2$.

However, be careful that this only holds as long as $r_1 \neq r_2$. For elementary volumes, this is always almost the case (except if you're looking for the probability of the two particles to be exactly at the same place), but when you are integrating this probability you need to be careful.

The joint probability formula is: $p(A \cup B) = p(A) + p(B) - p(A \cap B)$. Say you want to find the probability for having one of the particles inside a domain $\Omega_1$, and the other one in a domain $\Omega_2$ with $\Omega_1 \cap \Omega_2 \neq \varnothing$. If we call $A =$ "particle $1$ is in $\Omega_1$ and particle $2$ is in $\Omega_2$" and $B =$ "particle $1$ is in $\Omega_2$ and particle $2$ is in $\Omega_1$", then:

\begin{align} p(A \cup B) &= p(A) + p(B) - p(A \cap B)\\ &= \int_{\Omega_1, \Omega_2} \left|\psi(r_1,r_2)\right|^2 dr_1 dr_2 + \int_{\Omega_2, \Omega_1} \left|\psi(r_1,r_2)\right|^2 dr_1 dr_2 - \int_{(\Omega_1 \cap \Omega_2)^2} \left|\psi(r_1,r_2)\right|^2 dr_1 dr_2 \\ &= \int_{\Omega_1, \Omega_2} 2 \left|\psi(r_1,r_2)\right|^2 dr_1 dr_2 - \int_{(\Omega_1 \cap \Omega_2)^2} \left|\psi(r_1,r_2)\right|^2 dr_1 dr_2\\ \end{align}

Where the first two integrals are equal because of the symmetry properties of the wavefunction. The simplest "safety-check" is to take $\Omega_1 = \Omega_2 = \mathbb{R}$ (in 1D, or equal to the whole space otherwise). In that case, the probability of finding the particles anywhere is $1$. But because of the normalization of a 2-body wavefunction, $\int_{\mathbb{R}^2} 2 \left|\psi(r_1,r_2)\right|^2 dr_1 dr_2 = 2 > 1$ so we must obviously remove something to have a probability $\leq 1$.

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