# Why does a phonon obey the Bose statistic?

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.)

• wikipedia page says; phonons are bosons, since any number of identical excitations can be created by repeated application of the creation operator Dec 5 '17 at 13:15
• Just imagine what are consequences of phonons being fermions? At least they wouldn't have a classical limit. Dec 5 '17 at 13:28
• @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. Dec 5 '17 at 13:32
• 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. Dec 5 '17 at 14:05
• 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? Dec 5 '17 at 14:14