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Could somebody please explain why the phonon must be a Boson (strictly speaking, it must obey the Bose statistic) regardless what it is composed of? (As I have heard, the lattice vibration of both Bosonic and Fermionic system obey Bose statistics.)

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    $\begingroup$ wikipedia page says; phonons are bosons, since any number of identical excitations can be created by repeated application of the creation operator $\endgroup$ Dec 5 '17 at 13:15
  • $\begingroup$ Just imagine what are consequences of phonons being fermions? At least they wouldn't have a classical limit. $\endgroup$
    – Ice-Nine
    Dec 5 '17 at 13:28
  • $\begingroup$ @Ezk1t Could you clarify a bit more please? I quite don't get what you said "There wouldn't have a classical limit it the phonons are fermions". Thanks. $\endgroup$ Dec 5 '17 at 13:32
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    $\begingroup$ Quantization of a system in usual sense is a substitution of Poisson bracket with commutator This concept goes deeper, namely presenting second quantization formalism you can quantize boson or fermion operators, but for fermionic fields, for them to satisfy their statistics (exchange of particles grants minus sign) you would have to use anti-commutator $[A,B]_{+} = AB + BA$. And now look at this problem backwards, what if you want to go from operators to classical values? For bose particles you stumble on real numbers, and for fermions grassman algebra appear. $\endgroup$
    – Ice-Nine
    Dec 5 '17 at 14:05
  • $\begingroup$ How about other distribution function? For example, how could we know that the distribution function of the phonon is not Maxwell-Boltzmann or anything else? $\endgroup$ Dec 5 '17 at 14:14
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Lattice vibrations are mechanical ("sound") waves, so amount to collections of coupled harmonic oscillators. Their normal modes amount to individual oscillators with the usual continuous spectrum, and, upon ("second") quantization, discrete spectrum, linear in the integer oscillator excitation number, and unbounded above--until the lattice is torn apart. It does not matter what the constituents of the lattice are, fermions, bosons, or whatever. The positions of these constituents oscillate, and it is these motions which are quantized.

These quantum oscillators then are bosons.

If, instead, for some recondite reason, the spectrum of these modes only had occupation numbers 0 or 1, you'd call them fermions, but, as already hinted in the comments, it's hard to conceive of a classical mechanical collective motion that would quantize to fermions with such a ferociously limited spectrum. A classical limit is often associated with large occupation numbers, unavailable here, for each mode.

I have to confess I am out of my depth with topological fermions, however, and to what extent these exotic collective excitations are fermionic. Perhaps a condensed matter person could bring expertise to bear.

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