How to care about only one particle in a two-particle system

[snatchingthepi]

Homework Statement

Consider the symmetry and antisymmetric two-particle wave functions for a one-dimensional box with impenetrable walls at x = +- L/2. One particle occupies the ground state, and the other occupies the first excited state.

What is the probability to find a particle at position x for either case if we do not care about the position of the second particle.

Relevant Equations

See below

So for the 1D infinite well with the states above, I have

$\psi_{symmetric} = \frac{2}{L} [sin[\frac{\pi x_1}{L}]sin[\frac{2\pi x_2}{L}] + sin[\frac{2\pi x_1}{L}]sin[\frac{\pi x_2}{L}]]$
$\psi_{antisymmetric} = \frac{2}{L} [sin[\frac{\pi x_1}{L}]sin[\frac{2\pi x_2}{L}] - sin[\frac{2\pi x_1}{L}]sin[\frac{\pi x_2}{L}]]$

The question statement says to find the probability of finding a particle at a position $x$ for both cases if we "do not care about the position of the second particle". How do I do that? I thinking I might be able to simply integrate over the whole range for one particle, and then integrate from one edge of the well to the position x for the other? But I've never done anything like this and do not know.

 

[hutchphd]

I believe you are on the right path. You don't actually need to do the integrals because you know the single particle states are orthonormal (the integrals either 0 or 1). So write it out carefully and see if it makes sense. This is a useful exercise.

 

[TSny]

Note that in this problem the well extends from x = -L/2 to x = L/2. In this case, the ground state is not of the form sin(πx/L).

 

[snatchingthepi]

I *just* saw that! Thank you. I have now have legitimate answers that preserve normalization when I integrate over everything. Thank you!