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I am reading Quantum Mechanics from Griffiths (for the first time); in chapter 5 (2nd ed.) page-179 it is written:

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Now I really is not able to understand what Griffiths exactly wants to say by the marked lines! And one more question: are the particles apart from each other at any instant of time or can exist at the same place simultaneously for some "t". Any kind of Help is appreciated.

Note: I am using picture instead of Mathjax because at this moment I have not that much time to write those symbols and I am also not much fluent in that!

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2 Answers 2

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If we swap particles $a$ and $b$ so that $a$ is in state $\psi_a ( \mathbf{r}_2)$ and $b$ is in state $\psi_b (\mathbf{r}_1)$ then the system of two particles is now in state

$\psi_a (\mathbf{r}_2) \psi_b (\mathbf{r}_1)$

Since in general, $\psi_a$ and $\psi_b$ are different functions of $\mathbf{r}$, then

$\psi_a (\mathbf{r}_2) \psi_b (\mathbf{r}_1) \ne \psi_a (\mathbf{r}_1) \psi_b (\mathbf{r}_2)$

so we can see that by swapping $a$ and $b$ we have changed the state of the system. Griffiths' point is that this means we must be able to tell $a$ from $b$ - because if $a$ and $b$ were identical then swapping the particles would have no effect on the state of the system. If $a$ and $b$ were identical then $\psi(\mathbf{r}_1, \mathbf{r}_2)$ would have to be a symmetric function of $\mathbf{r}_1$ and $\mathbf{r}_2$, which is not generally true for $\psi_a (\mathbf{r}_1) \psi_b (\mathbf{r}_2)$.

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Allow me to use an analogy. Let me draw two imaginary chess boards. In the first, two pawns of the same color occupy opposite corners, whereas in the second one, it's one white and one black pawn.

In the second board, the pawns clearly have an "identity". One is black, the other is white. You will not mix them at any time. If they switch positions, you will know. But the first board is more complicated. My two white pawns are indistinguishable. They might be exchanging positions randomly without me ever noticing.

The situation described in the marked line is is the first board. There's really no point in trying to define pawn 1 and pawn 2, all I know is there are two pawns occupying corners.

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