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For 2 indistinguishable particles, we take the wave function to be $$\psi\pm (r_1,r_2) = A[\psi_a (r1)\psi_b (r2) \pm \psi_b (r1)\psi_a (r2) ]$$ where fermions get a - sign and bosons get a +

But, if both the particles are indistinguishable, isn't $\psi_a (r1) = \psi_a (r2)$, cause they behave the same way. Why do we take them to be different?

Also, can we calculate by some means that bosons get a + and fermions a -? or are they just experimental observations?

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  • $\begingroup$ What is the reasoning behind $\psi_a(r_1) = \psi_a(r_2)$, where $a$ labels a quantum number and $r_i$ labels the position of the $i^{th}$ particle? $\endgroup$ – JEB Dec 1 '18 at 19:26
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Indistinguishability means that if you swap particles of location, the description of the systems remains the same, up to a sign. So for example, if you have Bosons, then you get

$$ \psi_+(r_1, r_2) = \psi_+(r_2, r_1) $$

Note that this is quite different to say that $\psi_a(r_1) = \psi_a(r_2)$. This last statement only implies that the wave function describing a single particle is the same at location $r_1$ and location $r_2$

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The particles are indistinguishable, but that does not mean that their position is the same. If so, all the particles would be located in a single point of space.

I think the property of antisymmetry and symmetry (+ and -) of the bosons and fermions wave function is experimental.

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