Is the matrix $\sigma_{\mu\nu}$
$$\sigma_{\mu\nu} = \frac{i}{2} [\gamma_\mu, \gamma_\nu]$$
defined in FeynRules, FeynCalc or any similar packages? I know that $\gamma_\mu$ is Ga[mu]
in FeynRules.
The $\gamma$ matrices are built-in, but undocumented, as Internal`DiracGammaMatrix[]
. Their indexing is also a bit different from the wiki page:
Table[Internal`DiracGammaMatrix[k, "Basis" -> "Dirac"] // MatrixForm, {k, 4}]
$$\{\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix}\}$$
(Other possible settings for "Basis"
include "Chiral"
(the default) and "Majorana"
.)
Thus, you can implement the commutator like this:
Options[diracCommutator] = Options[Internal`DiracGammaMatrix];
diracCommutator[p_, q_, opts : OptionsPattern[]] :=
With[{pm = Internal`DiracGammaMatrix[p, opts], qm = Internal`DiracGammaMatrix[q, opts]},
I (pm.qm - qm.pm)/2]
and then you can do e.g.
diracCommutator[2, 4, "Basis" -> "Dirac"] // MatrixForm
$$\begin{pmatrix} 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & -i & 0 \\ \end{pmatrix}$$
In FeynCalc there is DiracSigma
DiracSigma[GA[mu], GA[nu]]
with
?DiracSigma
DiracSigma[a, b] stands for I/2*(a . b - b . a) in 4 dimensions. a and b must have Head DiracGamma, DiracMatrix or DiracSlash. Only antisymmetry is implemented.
However, you cannot have explicit Dirac indices attached to the Dirac matrices. This is something planned for the future (which is needed to use FeynCalc with QGraf)
The matrix $\sigma_{\mu\nu}$ is given by Sig[mu, nu]
in FeynRules, a related package, and defined
Sig[mu_,nu_,ss1_,ss2_]->I/2 TensDot[Ga[mu].Ga[nu]][ss1,ss2]-I/2TensDot[Ga[nu].Ga[mu]][ss1,ss2]},\[Infinity],Heads->True];
in Interfaces/FeynArtsInterface.m
.
$$
works, as you might now see. $\endgroup$