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In the treatment of Bosonic or Fermionic systems that I'm familiar with, you start with a state containing at least two particles:

$$ \left| a_{i}, a_{j} \right\rangle $$

And define a permutation operator, that switches their place. When you apply it twice to any state, you get back the same state.

$$ P_{ij}P_{ij}\left| a_{i}, a_{j} \right\rangle = P_{ij}\left |a_{j},a_{i} \right\rangle =\left | a_{i}, a_{j} \right\rangle $$

It is then argued that since elementary particles are indistinguishable from each-other, their composite states must be eigenstates of this operator, and the condition above constrains the eigenvalues to be either $1$, or $-1$, corresponding to totally symmetric and antisymmetric states, which gives us Bosons and Fermions.

What I'm having a hard time seeing is how this leads to such drastically different behavior at low temperatures. If we had a system of particles that were nearly identical but differed by some random mass $\delta m$ let's say, how would our partition function look? Would we recover Bose-Einstein or Fermi-Dirac statistics in the limit as $\delta m \rightarrow 0$ in a smooth way?

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Considering what happens to the partition function in the $\delta m \rightarrow 0$ limit is not really a sensible question for fermions and bosons, because "nearly identical" is nearly a quantum oxymoron. There is no smooth $\delta m \rightarrow 0$ limit in quantum mechanics that can convert non-identical particles into identical particles. If all their other intrinsic properties are the same, they are not identical if $|\delta m| > 0$, but they are identical if $\delta m = 0$. The transition between the two is discontinuous. This binary difference between identical and non-identical particles is fundamental in quantum mechanics, as is discussed in the answer to Almost identical fermions fighting for the same state.

I find it helpful to think about how two particles can possibly differ in quantum mechanics, and how mass differences are always associated with different quantum numbers. If two particles $A$ and $B$ are identical except that their rest masses differ, then $A$ and $B$ can simply be considered to be different states of the same particle, since there is nothing to prevent $B\leftrightarrow A$ transitions (e.g. by decay or collision or mixing). In any quantum theory describing $A$ and $B$, this means that state $B$ must have at least one quantum number whose value differs from state $A$, so they are not identical “particles” no matter how small $\delta m$ is.

For example, $A$ and $B$ could be hydrogen atoms in their $1S$ and $2S$ states. It is not just their tiny mass difference ($\sim0.000001\%$) that makes the $1S$ and $2S$ atoms distinguishable, but that their principle quantum numbers $(n=1,2)$ differ by $1$, corresponding to different internal quantum states and wave functions. Even if the electron was much lighter, making the mass difference between the two hydrogen states much smaller, their principle quantum numbers would still differ by $\Delta n = 1$.

Or consider the $1S$ hydrogen $F=0,1$ hyperfine states that differ in mass by less than $0.0000000000001\%$. Again the mass difference is tiny, but their angular momentum differs by $\Delta F = 1$, depending on whether the electron and proton spins are parallel or antiparallel.

Of course, explicit internal structure is not the only way mass differences can occur. For example, masses of fundamental particles in the Standard Model are generated by the Higgs Mechanism, and a popular explanation for why neutrinos are so light is the Seesaw Mechanism. What all these methods of generating mass share, however, is that different masses are always associated with different quantum states.

In a fully quantum system, otherwise identical quantum particles cannot differ only in mass. Any mass difference is associated with a difference in their intrinsic quantum properties that make them non-identical no matter how small the mass difference is. The transition from non-identical to identical particles requires changing the quantum state of the particles, which causes the dramatic changes in the partition function.

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  • $\begingroup$ I'm more concerned about the intrinsic indistinguishableness of the particles, and not differences in states of composite particles. Although, regarding your example with hydrogen, why must the principal quantum number be 1? Can't you have a superposition of 1S and 2S that will give a certain probability of n being 1 or 2? What's to stop me from supposing a large number of hydrogen atoms are nearly in the 1S state, and therefore nearly identical without a discontinuity? $\endgroup$ – Connor Dolan Nov 22 '17 at 20:38
  • $\begingroup$ @Connor Dolan I added a few more examples on how different masses require different quantum states, even if the particles are not composite. Having a superposition of 1S and 2S atoms with a small probability of being n a 2S state is very different from having "a large number of hydrogen atoms nearly in the 1S state". Such atoms would not be "nearly identical", since when the state of any individual atom was measured, it would be in either the 1S or 2S state. You are correct that for such a mixture there is no problematic discontinuity when the probability of being in a 2S state goes to zero. $\endgroup$ – David Bailey Nov 26 '17 at 15:03
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Two fermions cannot go in the same state because the permutation operator multiplies the state by -1 but also gives the same state, because the particles are identical. The only thing equal to its negative is zero.

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