# 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. Apr 3, 2020 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! Apr 3, 2020 at 16:10