# Is the Bogoliubov quasiparticle boson or fermion?

The Bogoliubov quasiparticle combines the properties of a negatively charged electron and a positively charged hole, so here we have two fermion and the quasiparticle have an integer spin. By this reason we have to consider this quasiparticle as boson, but in literature I came across with definition this one as fermion.Please, can anybody explain me what's wrong?

## 1 Answer

It depends.

Bogoliubov transformation has a fermion version $+$ and a boson verion $-$.

$$[c,c^\dagger]_{\pm}=1$$

The transformation is parameterized by $u$ and $v$ $$a=uc+vc^\dagger$$ $$a^\dagger=v^*c+u^*c$$

The goal is to restore similar commute relation as $c,c^\dagger$

$$[a,a^\dagger]_{\pm}=[uc+vc^\dagger,v^*c+u^*c]_{\pm}=(|u|^2\pm|v|^2)[c,c^\dagger]=1$$

The commutation relation determines the nature of fermion or boson.

Since you start with fermion, the Bogoliubov quasiparticle should also be fermion.

The "two" means a quantum superposition of two states, it does not mean you have two fermions particles.