Particle Indistinguishability Scale Limit

Question

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?

[1] http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf page 3.

Answer 1

As a technical term in quantum mechanics, two particles are indistinguishable if they cannot be distinguished even in principle. Typically, this means that they are identical other than their spin, momentum, position, energy and the time at which they exist. What this implies is that the wave state that describes the pair of particles has to be symmetrized or anti-symmetrized. In particular, different isotopes are distinguishable. One writes:
$$|AB\rangle = (|A\rangle|B\rangle \pm |B\rangle|A\rangle)/\sqrt{2}$$
where the sign is + for symmetrization of bosons, or - for the antisymmetrization of fermions. On the other hand, if the two states are not identical, then the wave state is not (anti)symmetrized:
$$|AB\rangle = |A\rangle|B\rangle$$
You can think of indistinguishability as arising naturally from the wave nature of particles. You might be able to write descriptive letters on particles, but how can you distinguish between two identical waves?

Now suppose you had a system where spin was conserved and never altered. An electron would be either spin-up or spin-down and never change and so you could distinguish between two such electrons. It would be as if they were distinct particles. This concept shows up in isospin where a neutron is treated as distinguishable from a proton even though they form an isospin doublet that is similar to the spin doublet (spin-down,spin-up). In such a restricted system, with a state of two particles you have the choice of writing your wave states either symmetrized, anti-symmetrized, or unsymmetrized. Any (consistent) way you choose, you'll get the same answer when you make your calculations as the two states form separate Hilbert spaces. However, if the instructor looks carefully at your derivation, it could be marked wrong even though it gives the correct answer -- it's better to follow the rules.

To learn more about this subtlety, read up on superselection sectors. What's going on is that when your Hilbert space divides into sectors which cannot physically interact, you do not have to symmetrize or antisymmetrize. But as I say above, you don't need to know this detail in order to calculate. Just follow the rules; only identical particles (in the sense of identical other than position, energy, momentum, orientation, etc.) are either symmetrized or antisymmetrized.