Does Fermi-Dirac Statistics explain anti-particles? I wondered whether the Fermi-Dirac Statistics describes the anti-fermion particles. Does it include the anti-particles?
 A: Antiparticles naturally arise when studying the Dirac equation within quantum field theory. Recall that we may expand a Dirac spinor field as a plane wave, namely,
$$\psi= \sum_{s=1}^2 \int \frac{\mathrm{d}^3 p}{(2\pi)^3} \frac{1}{\sqrt{2E_{p}}} \left[ b^s_p u^s(p)e^{ipx}+c^{s\dagger}_p v^s(p)e^{-ipx}\right]$$
and similarly for the conjugate field. Notice the appearance of two distinct creation and annihilation operators; these give rise to the electron and positron, the antiparticle.

The Dirac spinor transforms under a representation of the double cover of $SL(2,\mathbb{C})$ which is a reducible representation. Hence we may propose a decomposition or ansatz, 
$$\psi=u(p)e^{-ipx}$$
where $u(p)$ is a four-component Dirac spinor which may be broken down into a set of two-component spinors known as Weyl spinors (and with a reality condition, Majorana spinors):
$$u(p)=\left(
\begin{array}{c}
\sqrt{p\cdot \sigma}\, \xi\\
\sqrt{p \cdot \sigma}\, \xi\\
\end{array}
\right)$$
for $\xi^{\dagger}\xi=1$. The antiparticle, a positron, corresponds to a negative frequency solution, namely,
$$v(p)=\left(
\begin{array}{c}
\sqrt{p\cdot \sigma}\, \eta\\
\sqrt{p \cdot \sigma}\, \eta\\
\end{array}
\right)$$
where $\psi=v(p)e^{+ipx}$ instead. Notice both solutions have positive energy, as
$$E=\int \mathrm{d}^3 x \, T^{00}=\int \mathrm{d}^3 x \, \bar{\psi}(m-\gamma^i \partial_i)\psi \geq 0$$
(The above expression is obtained by applying Noether's theorem to the spacetime translation symmetry giving rise to energy-momentum tensor.)

Both the electron and positron are fermions, obey the same quantum field theory, and satisfy Fermi-Dirac statistics which - roughly - dictate we quantize the theory using anti-commutation relations rather than commutation relations, otherwise we would obtain a Hamiltonian unbounded from below.
