Multiplicity of distinguishable particles I have a problem in mind. There are 2 particles in a box. If they are distinguishable, there are 4 microstates represented by the picture below. If the particles are indistinguishable, there are only 3 microstates available (since the two states that are the lowest on the figure are the same state).
What I find strange is that if you want to count the probability to find 2 particles in the same half of the box at the same time, it is different for the two cases.
For the distinguishable case it is 1/2.
For the indistinguishable case it is 2/3.
If the particles are just moving there randomly, why should the system care if the particles are distinguishable or not. If we are to make measurements why should we see that in the indistinguishable case particles will occur at the same side of the box more often that in the distinguishable case?

 A: In your question, you have implicitly assumed that the particles are bosons, and not fermions. If the particles were fermions, the only allowed microstate would be for the two particles to be in separate boxes. Then the probability for the two particles to be in separate boxes would be 1.
What you are discovering is a "statistical force" involving indistinguishable particles. This appears in many other contexts in statistical mechanics. Roughly speaking, quantum statistics leads to an extra effective attraction between identical bosons, and effective repulsion between identical fermions. In your example, this effective attraction leads to a larger probability of finding two bosons in the same box, compared to the case with indistinguishable particles. On the other hand, for the fermion example, the effective repulsion leads to a smaller probability (in fact zero probability) of the two particles being in the same box.
I think one reason your question may be counter-intuitive is that, by working with 2 boxes, you have chosen an example in which quantum effects are, in some sense, very large. In particular, the effective bosonic attraction is very large. What your example essentially enforces by having so few allowed position states, is that the separation of the particles is never very large compared to the Compton wavelength of the particles.
We can see this more explicitly by generalizing your example, to consider 2 particles and $N$ boxes, which live on a line. Each particle can be in one of the $N$ boxes. I claim that the "classical limit" in which quantum effects are small, is the limit $N\rightarrow \infty$.

*

*For distinguishable particles, there are $N^2$ microstates. This is because are $N$ choices for where to place particle 1, and $N$ choices for where to place particle 2. Meanwhile, the probability $p$ that two particles will be in the same box is $1/N$. This is because are $N$ microstates where both particle 1 and 2 are in the same box, and $N^2$ total microstates, so $p=N/N^2=1/N$. As a sanity check, we recover your result $p=1/2$ when $N=2$. In the limit $N\rightarrow \infty$, where the two particles can explore a large space, the probability of a collision becomes vanishingly small: $p\rightarrow 0$.


*For indistinguisable bosons, there are $N(N+1)/2$ microstates. There are $N$ ways to choose the box for boson 1. For boson 2, there are $N-1$ ways to choose a box different from boson 1; however for each of these states we must divide by 2 to account for the indistinguishable nature of the bosons. Then there is 1 way2 to choose a box where the second boson is in the same box as the first. Putting this together, $N(N-1)/2+N=N(N+1)/2$. Meanwhile, there are $N$ microstatses where the 2 bosons are in the same state. We then have $p=2/(N+1)$. When $N=2$, we get $p=2/3$ as you computed. When $N\rightarrow \infty$, we have $p\rightarrow 0$, and therefore in the "classical limit" the effect of indistinguishability goes away. However, there is still some small residusal effect, since asymptotically for large $N$, $p \sim 2/N$ is a bit larger than the indistinguishable particle case, due to the statistical attraction.


*For indistinguishable fermions, there are $N(N-1)/2$ microstates. We have $N$ choices for where to put the first fermion. There are then $N-1$ choices for where to put the second fermion, since two fermions can't live in the same state. There are 0 states where the 2 fermions are in the same box. In the limit $N\rightarrow \infty$, we recover the limit obtained in the above two examples. Again there is still some residual statistical effect, since for large $N$, $p=0$, which is less than the distinguishable case, due to the statistical repulsion.
