How does the Pauli-exclusion principle work if space is infinitely divisible? Naively any two fermions should always be in different quantum states unless they are separated by an infinitesimally small distance and are in exactly the same position.

Is it possible that the Pauli-exclusion principle works because particles occupy a (De Broglie) wavelength and not just a point position in space?

Can two particles both have a probability of being in the same state but not have a probability of both being in the same state at the same time?

Is it possible that:

$$ \mathrm{P}(S(e_1) = X \, \vert \, S(e_2) \ne X) = \mathrm{P}(S(e_2) = X \, \vert \, S(e_1) \ne X)$$

but necessarily so that:

$$ \mathrm{P}(S(e_1) = S(e_2)) = 0$$

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    $\begingroup$ That is what quantization is about. the phase space is not infinitely divisible in bound systems. There are states with specific quantum numbers and two identical fermions cannot occupy quantized states with the same quantum numbers. $\endgroup$ – anna v Jul 3 '16 at 5:10
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    $\begingroup$ It's a lot cleaner if you start thinking about it in terms of field commutators since you aren't actually dealing with "particles" but with quanta of fields. $\endgroup$ – CuriousOne Jul 3 '16 at 5:58
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    $\begingroup$ In addition to what annav said, particles do not occupy a point position in space in quantum mechanics (except an idealized free particle). Bound states are always discrete. I'm not sure what your question is - you seem to be trying to apply the Pauli exclusion principle without actually knowing the states you want to apply it to. $\endgroup$ – ACuriousMind Jul 3 '16 at 10:18

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