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Recently when I deal with 1D electron system, it occurred to my mind that since these electrons are not able to bypass each other during the scattering processes, we can actually label them as the 1st, 2nd,..., Nth electron. As a result, it seems that these electrons now become distinguishable.

so my questions is: does this kind of distinguishability have any deep physical consequences? For instance, for 3D identical particles the wave function has to be either symmetric or anti-symmetric, whereas in 2D case we have the interesting anyons that obey a different statistics. Then what about the 1D case? Further more, what kind of distribution functions should we use (i.e. Fermi-Dirac or Bose-Einstein)? I do remember that in undergraduate condensed matter modules people deal with 1D electron gas using Fermi-Dirac distribution, but now it seems not so natural to me.

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One can also label the electrons in an atom by the energy in its Hartree-Fock approximation, and thus makes them distinguishable. This has physical consequences, for example one can speak unambiguously about the outer electron of a Lithium atom.

For a 1D quantum system one may have nonstandard statistics related to the braid group. Instead of Bose or Fermi statistics one has exchange relations satisfying Yang-Baxter equations. There is a nearly endless literature about this and the related quantum groups.

In 1D there is also no spin-statistic theorem, and one can describe bosons by fermionic fields and conversely.

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  • $\begingroup$ but I think the labelling of electrons in 1D case is fundamentally different from the label for electrons in an atom because it is possible for electrons in an atom to undergo transitions and thus come out with a different label, whereas in 1D case there is absolutely no way of doing that. $\endgroup$ – M. Zeng Dec 30 '15 at 15:35
  • $\begingroup$ @M.Zeng: A well-defined outer electron can make transitions (e.g., escape through ionization), or the complete electronic state can undergo transitions, but it doesn't make sense to speak of a transition swapping the second and third energy position. $\endgroup$ – Arnold Neumaier Dec 30 '15 at 15:46
  • $\begingroup$ yes, the levels cannot be swapped, but we are talking about electrons. If we leave alone the atom and let it undergo some drastic transitions, then we can no longer pinpoint the electron that used to be in the outer shell, which again is due to the indistinguishability. but in the 1d case, no matter how they interact they are ordered the same way they used to be. $\endgroup$ – M. Zeng Dec 30 '15 at 16:17
  • $\begingroup$ @M.Zeng: In 1D, when two particles touch and separate again you also cannot say which one was left of the other before. There are no particle position operators in a multiparticle system; thus particles are fiction (only the field made up by them exists) unless they can be uniquely labeled. But there is always one electron that is the outermost, and this one is distinguished. It doesn't become a different one since the notion of ''different'' is meaningless without the label. $\endgroup$ – Arnold Neumaier Dec 30 '15 at 16:26
  • $\begingroup$ so am I right to say that if we treat the two colliding electrons as the excitations or ripples of the underlying electron field, then it is possible for the two ripples to bypass each other just like what we have for two water ripples and thus indistinguishability still applies here? $\endgroup$ – M. Zeng Dec 31 '15 at 2:27
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Your intuition is exactly correct. In 1D, fermions and "hard-core" bosons (i.e. bosons with strong on-site repulsion that prohibits putting two bosons on the same site) are exactly dual to each other and produce the same energy spectrum for any given Hamiltonian. This (nonlocal) duality is easy to construct: a system of fermions is dual to a spin-1/2 chain by the (nonlocal) Jordan-Wigner transformation, and a system of hard-core bosons is dual to a spin-1/2 chain by the (local) Holstein-Primakoff transformation. By composing the two dualities together and "passing through" the intermediate spin-1/2 chain, you get get a duality between fermions and hard-core bosons.

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