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In Bose-Einstein statistics, two identical bosons can be in one of the following three states with respect to some orthonormal single particle states $|A\rangle$ and $|B\rangle$:

(1) $|A\rangle |A\rangle$

(2) $|B\rangle |B\rangle$

(3) $1/\sqrt{2}(|A\rangle|B\rangle + |B\rangle|A\rangle)$

Now, (3) is a linear superposition of two states, and we would never really observe this superposition state. If the system is measured, I suppose, (3) will collapse into either $|A\rangle |B\rangle$ or $|B\rangle|A\rangle$ with the probability of 1/2.

First, is this right? If it is right, what would be the observable for which $|A\rangle |B\rangle$ and $|B\rangle|A\rangle$ are both eigenstates? In other words, which observable is associated with the measurement that can collapse (3) into $|A\rangle |B\rangle$ or $|B\rangle|A\rangle$?

Secondly, and relatedly, it looks like $|A\rangle |B\rangle$ and $|B\rangle|A\rangle$ should not be allowed as possible physical states, for they are not permutation invariant. But it also looks like (3) should collapse into one of those states if measured. I'm confused. Please enlighten me. Thanks.

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As you can see, the system has three distinct states. You should read the configuration (3) as "the probability of finding the first particle in the $\mid A \rangle$ state with the second one being in the state $\mid B \rangle$ (and vice-versa)". So each one of these three states has an equal probability $\frac{1}{3}$. As for "what would be the observable?", we are talking about energy aren't we? In BE statistics, we are concerned with how bosons occupy certain energy states.

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    $\begingroup$ Thanks for your answer. Do you mean that (3) should be read as "the probability of finding the first particle in A with the second one being in B OR the first in B with the second in A"? But doesn't this entail that it is a possible experimental outcome that the first particle is in A and the second in B, which violates permutation invariance? The whole point of (3), as I understand it, is that AB or BA is not a physically meaningful state becasue it is not permutation invariant. If you symmetrize AB or BA, you get (3). But if (3) entails a non-symmetric result, something seems wrong $\endgroup$ Sep 2 '20 at 2:34

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