I wondered whether the Fermi-Dirac Statistics describes the anti-fermion particles. Does it include the anti-particles?

  • $\begingroup$ Particles and their anti-particles both should have the same statistics -- so both bosonic or both fermionic. Bosonic examples: $\pi^{\pm}$ or $W^{\pm}$. $\endgroup$
    – Siva
    Apr 29, 2014 at 16:15

1 Answer 1


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,


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.

  • $\begingroup$ So, if I remember the derivations correctly, we can say that from Dirac's equation we directly get both antiparticles and the statistics. Right? $\endgroup$
    – Davidmh
    Apr 29, 2014 at 19:50
  • $\begingroup$ @Davidmh: What do you mean by "get statistics from the Dirac equation"? $\endgroup$
    – JamalS
    Apr 29, 2014 at 19:52
  • $\begingroup$ @Davidmh: The Fermi-Dirac distribution is derived using a statistical ensemble. In a way quantizing using anti-commutations arises because commutation relations are inadequate, and hence they are the only reasonable canonical quantization alternative. $\endgroup$
    – JamalS
    Apr 29, 2014 at 19:54

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