# Calculating One-Dimensional Particle Separation Probabilty Density

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?

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.
• No problem! And in fact you do account for that -- it's just $P(0)$ in that formula. – mthibodeau Apr 3 '20 at 15:32
• Put dollar signs around your formula to make it render inline (like $x = 2$) or two dollar signs on each side to put the formula on its own line. Also if you do like the answer don't forgot to mark it as accepted! – mthibodeau Apr 3 '20 at 16:10