2 fermions in a box (infinite potential well) I have 2 fermions in a box.
I know that they are in the state:
$$|\psi\rangle = {1 \over \sqrt2}\, 
(|1\rangle |2\rangle -|2\rangle|1\rangle)\,|+,+\rangle$$
If I hadn't spin, I could find wave function of the state doing:
$\langle x|\psi\rangle$,
knowing that eingenfunctions of the single particle are :
$${\sqrt {2\over L}}\,\sin \left(n \pi {x \over l}\right).$$
So if I want to calculate eingenfunction for the state with spin, is there something I can do?
 A: Your $|\psi\rangle$ is the wave function, apart from containing a spatial part it just also consists of a spin part. What you'd like to calculate - $\langle x|\psi\rangle$ - is the spatial component of the wave function alone. In order to obtain this, we need to trace out the spin part (i.e. forget about it).
This is complicated by $|\psi\rangle$ being a symmetrized two-particle wave function. There are two particles which each have a degree of freedom ($|x_1\rangle$ and $|x_2\rangle$). However, if we were to calculate $\langle x_1|\psi\rangle$ we wouldn't necessarily get something that made much sense:
$$
  \langle x_1|\psi\rangle = \frac{1}{\sqrt{2}} (f_1(x_1)|2\rangle - f_2(x_1)|1\rangle)|+,+\rangle,
$$
$f_n(x)$ being the single-particle wave function in a box potential.
Since these are indistinguishable particles you can't simply calculate $\langle x_i|\psi\rangle$ anymore, and expect something that resembles a probability amplitude.
You could project onto a state that gives you something resembling the total amplitude for finding a particle (any particle) at the position $x$ - but I'm not sure whether that's what you're after. This would look like:
$$
  \left(|x\rangle \otimes \sum_n |n\rangle - \sum_n |n\rangle \otimes |x\rangle\right) \otimes \sum_{s_1,s_2} |s_1, s_2\rangle.
$$
Here the tensor products ($\otimes$) emphasize that we're dealing with multiple bases (one for each spatial degree of freedom of the box potential, and one for the spins). By summing over them like this we trace out these degrees of freedom, leaving a result that resembles a probability amplitude, and which is proportional to $f_1(x) - f_2(x)$.
