Fermion field

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Fermions are particles whose quantum mechanical wavefunction is totally antisymmetric under quantum number interchange. We will denote the Fermion (or Dirac) field by <math>\psi(x)</math>. This requirement imposes the anticommutation rules

<math>\{a^{r}_{\textbf{p}},a^{s \dagger}_{\textbf{q}}\} = \{b^{r}_{\textbf{p}},b^{s \dagger}_{\textbf{q}}\}=(2 \pi)^{3} \delta^{3} (\textbf{p}-\textbf{q}) \delta_{ab}.\,</math>

where <math>a^{s \dagger}_{\textbf{p}}</math> creates a fermion of momentum <math>\textbf{p}</math> and spin <math> s </math>, and <math>b^{s \dagger}_{\textbf{p}}</math> creates an antifermion. These anticommutation relations require that the plane wave expansion for <math> \psi(x) </math> create an antifermion while annhilating a fermion

<math>\psi(x) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2E_{p}}}\sum_{s} \left(

a^{s}_{\textbf{p}}u^{s}(p)e^{-ip \cdot x}+b^{s \dagger}_{\textbf{p}}v^{s}(p)e^{ip \cdot x}\right).\,</math>

From this it follows that the equal-time anticommutation relations for the field operators are

<math>\{\psi(\textbf{x}),\psi^{\dagger}(\textbf{x})\} = \delta^{(3)}(\textbf{x}-\textbf{y})</math>

Since all observables are built out of an even number of fermion fields, the commutation relation vanishes between any two observables at spacetime points within the light cone. As we know from elementary quantum mechanics two simultaneously commuting observables cannot be measured simultaneously. We have therefore correctly implemented Lorentz invariance for the Fermion field, thereby preserving causality.

It can be straightforwardly verified that our expression for <math>\psi(x)</math> satisfies the Dirac equation

<math>(i\gamma^{\mu} \partial_{\mu} - m) \psi(x) = 0.\,</math>

Note that the Dirac equation can be derived from the Fermion Lagrangian

<math>\mathcal{L}_{Fermion} = \bar{\psi}(i\gamma^{\mu} \partial_{\mu} - m)\psi.\,</math>

where "psi-bar" is defined as <math>\bar{\psi} \equiv \psi^{\dagger} \gamma^{0}</math> .

Given the expression for <math>\psi(x)</math> we can construct the Feynman propagator for the Fermion field

<math> D_{F}(x-y) = \langle 0| T(\psi(x) \bar{\psi}(y))| 0 \rangle </math>

we define the time-ordered product for Fermion with a minus sign due to their anticommuting nature

<math> \langle 0| T(\psi(x) \bar{\psi}(y))| 0 \rangle \equiv \theta(x^{0}-y^{0}) \langle 0| \psi(x) \bar{\psi}(y)| 0 \rangle - \theta(y^{0}-x^{0}) \langle 0| \bar{\psi(y)} \psi(x)| 0 \rangle .\,</math>

Plugging our plane wave expansion for the Fermion field into the above equation yields:

<math> D_{F}(x-y) = \int \frac{d^{4}p}{(2\pi)^{4}} \frac{i(p\!\!\!/ + m)}{p^{2}-m^{2}+i \epsilon}e^{-ip \cdot (x-y)}</math>

Where we have employed the Feynman slash notation. This result makes perfect sense since the factor

<math>\frac{i(p\!\!\!/ + m)}{p^{2}-m^{2}}</math>

is just the inverse of the operator acting on <math>\psi(x)</math> in the Dirac equation. Note that the Feynman propagator for the Klein-Gordon field has this same property.

The basics of free Fermion field theory are sketched here. For more complicated field theories involving interactions (such as Yukawa theory, or quantum electrodynamics), the results are known only perturbatively.