Decomposition of two particle wavefunction into product of single-particle wavefunctions Suppose you prepare a two-particle system such that $\Psi(\vec{r}_1,\vec{r}_2, t_0) = \Psi_1(\vec{r}_1, t_0)\Psi_2(\vec{r}_2, t_0)$.
So then, initially $\Psi(\vec{r}_1,\vec{r}_2,t_0) - \Psi_1(\vec{r}_1, t_0)\Psi_2(\vec{r}_2, t_0) = 0$
But now you let the system evolve in time and $\Psi(\vec{r}_1,\vec{r}_2, t)$ can now no longer be decomposed into a product of two single-particle wavefunctions.
What then does, physically, does $\Psi(\vec{r}_1,\vec{r}_2, t) - \Psi_1(\vec{r}_1, t)\Psi_2(\vec{r}_2, t)$ represent?
 A: You might be interested by the correlation between two observables relatives to particles $1$ and $2$, for instance $X_1$ and $X_2$ :  
$corr(X_1,X_2) = \dfrac{cov(X_1, X_2)}{\sigma_{X_1}\sigma_{X_2}} = \dfrac{\langle X_1X_2  \rangle - \langle X_1  \rangle \langle X_2  \rangle }{\sqrt{\langle X_1^2  \rangle \langle X_2^2  \rangle}} \tag{1}$
The covariance $cov(X_1,X_2)$ may be written : 
$cov(X_1,X_2) = \langle X_1X_2  \rangle - \langle X_1  \rangle \langle X_2  \rangle \\=\int d^3  \vec r_1 d^3 \vec r_2 ~ (|\psi(\vec r_1,\vec r_2)|^2 - |\psi_1( \vec r_1)|^2|\psi_2(\vec r_2)|^2) ~(x_1x_2)\tag{2}$
When the particles $1$ and $2$ are independent, you have the factorization $\psi(\vec r_1,\vec r_2) = \psi_1(\vec r_1) \psi_2(\vec r_2)$, so, all the correlations between observables relative to particles $1$ and $2$ are zero, as wished.
So, the expression $(|\psi(\vec r_1,\vec r_2)|^2 - |\psi_1( \vec r_1)|^2|\psi_2(\vec r_2)|^2)$ might be seen as a general (observable-independent) measure of correlations (more precisely covariance) at the point ($\vec r_1, \vec r_2$)
