2
$\begingroup$

Suppose I have a box divided in a left and a right side and two spinless identical particles moving at random in the box. I measure their position and I am interested in how many particles are on the left and on the right side of the box. According to classical statistical mechanics I should have a probability of $ \frac{1}{4} $ to have both particles on the right side of the box, a probability $ \frac{1}{4} $ to have them on the left side and a probability of $ \frac{1}{2} $ to have one particle on each side. That is beacause I have a probability $ \frac{1}{4} $ of having the first particle on the right side of the box and the second one on the left side and another $ \frac{1}{4} $ of it being vice versa. I expect it to be the right answer. If I see the same problem from a quantum mechanical point of view I get the wrong (?) answer and I would like to understand why is my reasoning flawed. Since particles are indistinguishable in the deepest sense I cannot speak of first and second particle, therefore I should have a probability $ \frac{1}{3} $ of both particles being on the right side, a probability of $ \frac{1}{3} $ of both particles being on the left side and a probability $ \frac{1}{3} $ of having one particle on each side. I tried to visualize it in terms of wavefunctions and it just doesn't solve the issue to me (in fact, it just makes it worse).

If I apply the same reasoning to a huge number of particles this indistinguishability seems catastrophic. If I am correct it would imply that the state where all particles are on the left side has the same probability of the state where particles are split half and half between the two sides!

$\endgroup$
0

1 Answer 1

3
$\begingroup$

The paradox is resolved by remembering to account for the various possible locations of a particle within a side. This is important because the allowed locations are all distinguishable, even though the particles are indistinguishable.

To count states, we either need to restrict the number of locations in each side to be finite (but arbitrarily large), or we need to treat the particles as truly quantum (not just indistinguishable) so that we can count mutually orthogonal wavefunctions. The first approach is sufficient for resolving the paradox, so I'll use it.

Let $K$ be the number of distinct locations in each side. For two particles, the number of states with both particles on the left side is $(K+1)K/2$, and the number of states with the particles on different sides is $K^2$.

With four particles, the number of states with all four on the left side is $\sim K^4/24$ for large $K$, while the number of states with two on the left and two on the right is $\sim (K^2/2)^2$, approximately six times larger than the number of all-on-left states.

For $2N$ of particles, the number of all-on-left states is $\sim K^{2N}/(2N)!$ for large $K$, and the number of half-and-half states is $\sim (K^{N}/N!)^2$. As $K\to\infty$, the ratio approaches $(2N)!/(N!)^2$, which is a rapidly increasing function of $N$. For $N=10$, it is already $> 10^5$. So, for a large number of indistinguishable particles in a macroscopic box, the number of half-and-half states greatly exceeds the number of same-side states.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.