# Particle Indistinguishability Scale Limit

QFT says that all particles are indistinguishable from one-another [1]. That is, take a proton from cosmic ray from a supernova a billion light-yeas away, and compare it to a proton that just got smashed out in the LHC, and they are indistinguishable from one-another. Replace either with the other, and nothing will change. (As I understand) this applies to all particles that arise from the standard model fields- fermions and bosons. So if I have an atom, all the electrons in the atom are indistinguishable from all the other electrons in the atom. Theoretically I then take that atom and interchange it with any other "indistinguishable" atom. Or can I?

Obviously I can't replace hydrogen-1 with deuterium (their masses are different), so atoms are at least distinguishable to their isotopes. What about within isotopes? The volume of an orbital in an electron cloud is larger than the Compton length of the electron by a factor of about 10^7, giving a tremendous volume that is vacuum, and the probability of getting the election clouds just right to have negligible effects on the quantum field seems unlikely. So, are two ${}^2\text{H}$ atoms indistinguishable from one another?

Let's say that does work, and we can say isotopes are indistinguishable. Now, I pose the same question for molecules. What do I need to know about two water molecules? In QFT and GR, there are the 4 spacetime dimensions. I clearly can't just take a water and replace it with another water by swapping their \$(\vec{x}, t) properties, because they might be "upside -down" and then all the hydrogen bonds will get messed up and their dipole moments will be wacky (and let's not even think about ionic or superionic water) and there will be some change in the system while the molecule "rights" itself. So, it seems I have at least answered my question at that scale - molecules are not indistinguishable from one another on their physical properties alone. For polar molecules in a "normal" form, we have at least 6 dimensions that must be swapped (assuming we can get away with calling the atoms themselves indistinguishable).

The final question, then, is what is the limit that systems are no longer indistinguishable based on just their physical properties?

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deuterium does not only have a different mass. Leaving out quarks for now, one is a system of three particles (proton, neutron, electron) the other of just two particles (proton, electron). –  luksen Aug 12 '11 at 17:21
Which is exactly part of my point with saying atoms are not possibly indistinguishable, you'd have to go at best to isotopes. –  David Souther Aug 12 '11 at 17:25

$$|AB\rangle = (|A\rangle|B\rangle \pm |B\rangle|A\rangle)/\sqrt{2}$$
$$|AB\rangle = |A\rangle|B\rangle$$