# Indistinguishable particles and probability density

I am given the following (probably simple) exercise, but I think I misunderstand something:

Let $\psi_{a,b}(r_1,r_2)$ be a two-particle state, calculate the probability density for distinguishable and indistinguishable particles.

I don't quite understand what this exercise wants me to do: If $\psi_{a,b}(r_1,r_2)$ , then this density should be given by $|\ \psi_{a,b}(r_1,r_2)|^2$, if our particles are indistinguishable, then I am not sure whether I am supposed to symmetrize or antisymmetrize ( or do anything else to this state) the wavefunction.

I mean, somehow this exercise wants me to get two different results for the cases, but I don't see which states exactly.

• Where did you get this assignment? It seems very poorly formulated. If it is from a teacher, I would ask him probability of what is the demanded density supposed to quantify. – Ján Lalinský Nov 6 '14 at 14:14
• You can get something more here physics.stackexchange.com/q/37556/5787 – Abinash Chakraborty Jun 22 '15 at 15:43

The probability density to find one particle at $r_1$ and the other particle at $r_2$ is the absolute square of the wave-function. The question is, so I understand, how does the wave-function look like. If the particles a and b are distinguishable, the wave-function looks as given in the exercise. However, if the particles are indistinguishable fermions, the wave function should be antisymmetrical at the interchange of the particles, i.e. you do antisymmetrization:
$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) - ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$
$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) + ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$