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Suppose I have some fermions with spin 1/2 on a harmonic potential. Then the energy of each particle is given by:

$$ E_i=\hbar\omega(n_{x_i}+n_{y_i}+n_{z_i}+3/2) $$

By definition the partition function is:

$$ Z=\sum_{\{\vec\mu\}}\exp(-\beta E(\vec\mu)) $$

where $\vec\mu$ is the microstate of the system. Being fermions I know that both can't possibly be in the same quantum state at once. Then the sums would be something like $\sum_{n_{x_i}\neq n_{x_j}}$, but my major doubt comes on the accounts of distinguishability, because those particles are only distinguishable if their spin quantum number, $m_s$, is the same for both particles, but if that isn't the case then the particles are in fact distinguishable and than I am not even sure that the fact that they can't be in the same quantum state would be significant because having different quantum spin numbers, it is impossible for them to be in the same quantum state, as they belong to different parts of the joint Hilbert space. How does one account for this?

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  • $\begingroup$ why are they distinguishable? $\endgroup$
    – Superfast Jellyfish
    Commented Jan 28, 2020 at 7:31
  • $\begingroup$ Because of their spin. If I know before hand that one particle has spin ul and the other spin down, then they are clearly distinguishable $\endgroup$
    – Bidon
    Commented Jan 28, 2020 at 10:39
  • $\begingroup$ How do you keep track of the up electron? $\endgroup$
    – Superfast Jellyfish
    Commented Jan 28, 2020 at 10:44
  • $\begingroup$ Suppose I decided to collide the two electrons. Then one would go in one direction and the other would go in other direction. Having different spins, I could for example set them up via stern gerlach experiments before and after the collision, keeping track of the particles $\endgroup$
    – Bidon
    Commented Jan 28, 2020 at 10:47
  • $\begingroup$ Yes. But we can’t distinguish the electrons after the collision until we make a measurement $\endgroup$
    – Superfast Jellyfish
    Commented Jan 28, 2020 at 10:49

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Particles are not distinguishable or indistinguishable based on their state but rather whether they are excitation of the same field. When we not want to involve this level of abstraction, we simply state (or given) if particles are distinguishable or not. Two particles of the same type will be indistinguishable, in general.

So the question of whether your two fermions are distinguishable or not is not related to the projection of their spin or to the energy level they occupy. These are details of the state. Rather, if the particles are identical (that is, both are electrons, or neutrons, or muons etc.) then we will treat them as indistinguishable.

Tow indistinguishable fermions really cannot occupy the same state. This is because we require that their full wave-function will be anti-symmetric under their exchange. Here the state they occupy come into play. If both of them have the same spin projection (identical $m_z$) then their spatial wave function must be antisymmetric. If, on the other hand, their spin projection is anti-symmetric, then their spatial wave function must be symmetric! (so the total wave function will be anti-symmetric under exchange).

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  • $\begingroup$ Could you explore a bit more on the first paragraph? I feel that my confusion is arising on the fundamental definition of we call distinguishable in quantum mechanics $\endgroup$
    – Bidon
    Commented Jan 28, 2020 at 10:56
  • $\begingroup$ particles in quantum theory are excitations of fields. The entire issue of spin and statistics of particles (that is if they obey Bose-Einstein or Fermi-Dirac statistics) is derived from the nature of the fields. However, for many applications we do not need to use the full abstraction and complex mechanism of quantum field theory. We do need to take care of the statistics of the particles. So just as we take other properties of them as given description (their spin, charge, etc.) we also define whether they are distinguishable or not. $\endgroup$
    – user245141
    Commented Jan 28, 2020 at 11:03
  • $\begingroup$ But isn't the whole idea of indistuishability that, if on a collision experiment, for indistinguishable particles, there's was no experiment that we could possibly envisage that would keep track of the particles whereas with distinguishable we can? Therefore with different spins they are distinguishable because we could design some experiment to keep track of the spins $\endgroup$
    – Bidon
    Commented Jan 28, 2020 at 11:05
  • $\begingroup$ I am not familiar with this definition of indistinguishability. For collision you need an interaction term - what is its nature? I think of it in the following term - we can do a subset of manipulations of the coordinate system: rotation, for example, or boosting. If using this we can map one particle onto the another, then they are indistinguishable. Here, if we rotate the spin about the $x$-axis by $\pi$, the up-spin will be a down-spin. However, no matter how we rotate or boost or whatever, the total spin of each electron will always be $1/2$. $\endgroup$
    – user245141
    Commented Jan 28, 2020 at 11:12

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