Bilinear Covariants

We have seen that the constant \bgroup\color{black}$\gamma$\egroup matrices can be used to make a conserved vector current

\begin{displaymath}\bgroup\color{black} j_\mu=ic\bar{\psi}\gamma_\mu\psi \egroup\end{displaymath}

that transforms correctly under Lorentz transformations. With 4 by 4 matrices, we should be able to make up to 16 components. The vector above represents 4 of those.

The Dirac spinor is transformed by the matrix \bgroup\color{black}$S$\egroup.

\begin{displaymath}\bgroup\color{black} \psi'=S\psi \egroup\end{displaymath}

This implies that \bgroup\color{black}$\bar{\psi}=\psi^\dagger\gamma_4$\egroup transforms according to the equation.

\begin{displaymath}\bgroup\color{black} \bar{\psi}'=(S\psi)^\dagger\gamma_4=\psi^\dagger S^\dagger\gamma_4 \egroup\end{displaymath}

Looking at the two transformations, we can write the inverse transformation.

\begin{eqnarray*}
S_{rot}=\cos{\theta\over 2}+\gamma_i\gamma_j\sin{\theta\over 2...
...\dagger\gamma_4=\psi^\dagger \gamma_4S^{-1}=\bar{\psi}S^{-1} \\
\end{eqnarray*}


This also holds for \bgroup\color{black}$S_P$\egroup.

\begin{eqnarray*}
S_P=\gamma_4 \\
S_P^\dagger =\gamma_4 \\
S_P^{-1}=\gamma_4 \...
...P^\dagger\gamma_4=\gamma_4\gamma_4\gamma_4=\gamma_4=S_P^{-1} \\
\end{eqnarray*}


From this we can quickly get that \bgroup\color{black}$\bar{\psi}\psi$\egroup is invariant under Lorentz transformations and hence is a scalar.

\begin{displaymath}\bgroup\color{black} \bar{\psi}'\psi'=\bar{\psi}S^{-1}S\psi=\bar{\psi}\psi \egroup\end{displaymath}

Repeating the argument for \bgroup\color{black}$\bar{\psi}\gamma_\mu\psi$\egroup we have

\begin{displaymath}\bgroup\color{black} \bar{\psi}'\gamma_\mu\psi'=\bar{\psi}S^{-1}\gamma_\mu S\psi=a_{\mu\nu}\bar{\psi}\gamma_\nu\psi \egroup\end{displaymath}

according to our derivation of the transformations \bgroup\color{black}$S$\egroup. Under the parity transformation

\begin{displaymath}\bgroup\color{black} \bar{\psi}'\gamma_\mu\psi'=\bar{\psi}S^{...
...amma_\mu S\psi=\bar{\psi}\gamma_4\gamma_\mu\gamma_4\psi \egroup\end{displaymath}

the spacial components of the vector change sign and the fourth component doesn't. It transforms like a Lorentz vector under parity.

Similarly, for \bgroup\color{black}$\mu\neq\nu$\egroup,

\begin{displaymath}\bgroup\color{black} \bar{\psi}\sigma_{\mu\nu}\psi\equiv\bar{\psi}i\gamma_\mu\gamma_\nu\psi \egroup\end{displaymath}

forms a rank 2 (antisymmetric) tensor.

We now have 1+4+6 components for the scalar, vector and rank 2 antisymmetric tensor. To get an axial vector and a pseudoscalar, we define the product of all gamma matrices.

\begin{displaymath}\bgroup\color{black} \gamma_5=\gamma_1\gamma_2\gamma_3\gamma_4 \egroup\end{displaymath}

which obviously anticommutes with all the gamma matrices.

\begin{displaymath}\bgroup\color{black} \{\gamma_\mu,\gamma_5\}=0 \egroup\end{displaymath}

For rotations and boosts, \bgroup\color{black}$\gamma_5$\egroup commutes with \bgroup\color{black}$S$\egroup since it commutes with the pair of gamma matrices. For a parity inversion, it anticommutes with \bgroup\color{black}$S_P=\gamma_4$\egroup. Therefore its easy to show that \bgroup\color{black}$\bar{\psi}\gamma_5\psi$\egroup transforms like a pseudoscalar and \bgroup\color{black}$\bar{\psi}i\gamma_5\gamma_\mu\psi$\egroup transforms like an axial vector. This now brings our total to 16 components of bilinear (in the spinor) covariants. Note that things like \bgroup\color{black}$\gamma_5\sigma_{12}=i\gamma_1\gamma_2\gamma_3\gamma_4\gamma_1\gamma_2=-i\gamma_3\gamma_4$\egroup is just a constant times another antisymmetric tensor element, so its nothing new.

Classification Covariant Form no. of Components
Scalar \bgroup\color{black}$\bar{\psi}\psi$\egroup 1
Pseudoscalar \bgroup\color{black}$\bar{\psi}\gamma_5\psi$\egroup 1
Vector \bgroup\color{black}$\bar{\psi}\gamma_\mu\psi$\egroup 4
Axial Vector \bgroup\color{black}$\bar{\psi}\gamma_5\gamma_\mu\psi$\egroup 4
Rank 2 antisymmetric tensor \bgroup\color{black}$\bar{\psi}\sigma_{\mu\nu}\psi$\egroup 6
Total 16
The \bgroup\color{black}$\gamma$\egroup matrices can be used along with Dirac spinors to make a Lorentz scalar, pseudoscalar, vector, axial vector and rank 2 tensor. This is the complete set of covariants, which of course could be used together to make up Lagrangians for physical quantities. All sixteen quantities defined satisfy \bgroup\color{black}$\Gamma^2=1$\egroup.

Jim Branson 2013-04-22