I think when two particles (with associated wavefunctions) with wavefunctions $\psi_1\left(x\right)$ and $\psi_2\left(x\right)$ overlap (i.e. they interact in vacuum) they will interfere and the resultant wavefunction will be $\psi\left(x\right)=\psi_1\left(x\right)+\psi_2\left(x\right)$ with some normallization constant. If there is constant phase difference between two wavefunctions, i.e. two wavefunctions are coherent with each other you will see time stationary fringes.
On the other hand when two particles interact in a medium then the medium's response can use both wavefunctions in a certain manner and you might see the multiplication of two in certain manner.
For example if the medium is linear then the final wavefunction will be addition $\psi\left(x\right)=\psi_1\left(x\right)+\psi_2\left(x\right)$
If the medium possesses second order nonlinearity i.e. $\psi\left(x\right)=\left(\psi_1\left(x\right)+\psi_2\left(x\right)\right)^2$, then you will see the final wavefunction having multiplication of the two wavefunctions. If two wavefunctions have frequencies $\omega_1$ and $\omega_2$ you will see the wavefunction with frequency component $\omega_1\pm\omega_2$, you will see higher powers of product if the medium has larger degree of nonlinearity.
If you still have some doubt please leave a comment.
@anon0909 : You have misunderstood the answer. If two particles occupy states $\psi_1\left(x\right)$ and $\psi_2\left(x\right)$ individually then it does not mean that you can distinguish between them when they interact.
Question is how you know about particles's state and the answer is via. measurement. When you measure the state of a particle you will get
$\left|\psi_1\left(x\right)\right|^2$ and $\left|\psi_2\left(x\right)\right|^2$,
when you make measurement after the interaction you will get
In this way both particle remain indistinguishable. Moreover if you make measurement before the interaction and try to interact these particles after the measurement was made you will just get $\left|\psi_1\left(x\right)\right|^2$ + $\left|\psi_2\left(x\right)\right|^2$ and no interference will be seen.
You may think my explanation as double slit interference experiment (although this can be applied to several other physical situations)
My effort was to put the tedious concepts in a simple manner rather than to stress over mathematical rigour.