Calculating One-Dimensional Particle Separation Probabilty Density

Question

Question

Today I am inquiring how one would calculate the particle separation probability density for 2 particles in a square well, for 2 distinguishable particles. We are given both particles wave functions as $\psi_1(x)$ and $\psi_2(x)$ with the probability separation density as $P(x_1-x_2)$. My biggest issue seems to be that I don't know how to create a wave function that is a difference between the two other wave functions.

My attempt at a solution

For a conventional wave function, we can calculate the spatial wave function probability density as:

$$P(x) = |\psi(x)|^2$$

Our problem wants us to calculate:

$$P(x_1-x_2)$$

which I interpret as some function of both particles, $\Psi(x_1,x_2)$, not the difference of the two functions. How can we create such a function?

Answer 1

Suppose we have a square well of width $a$. If $s = x_1 - x_2$ is the separation, the density you want is given by $$P(s) = \int_0^{a-s} dx \,\left(|\psi_1(x)|^2|\psi_2(x+s)|^2+|\psi_1(x+s)|^2|\psi_2(x)|^2\right)$$ Where did this come from? You want the joint probability that you'll find particle 1 at position $x$ and particle 2 and a distance $s$ away from it (i.e. at position $x + s$), which is given by the first term in the integral $|\psi_1(x)|^2|\psi_2(x+s)|^2$. You need to also account for having particle 2 at $x$ and particle $1$ at $x+s$, so you have to include the second term in the integral as well. Since you don't care what your base point $x$ is, integrating over all $x$ in the domain gives you the answer you want.