The terms (pseudo)scalar, (axial)vector and tensor represent the structure of a given meson in Dirac space. So, $\bar\psi\Gamma_i\psi$ with $i\in$ {scalar, pseudoscalar, ...}.
Any $4\times 4$ spinor matrix $\Gamma$ can be decomposed in a linear combination of the unit matrix $\textbf 1$ or products of $\gamma$-matrices and thus into terms with well-defined Lorentz transformation properties when sandwiched by Dirac fields: $\psi\Gamma\psi$. This is then called a $\textit{Dirac bilinear}$.
$\textbf{Lorentz transformation properties.}$ How something transforms under a Lorentz transformation is determined by its free Minkowski indices. A scalar transforms like, $$U^{-1}S(x)U = S(\Lambda ^{-1}x),$$ regardless how this operator $U$ looks in detail (this is determined by the nature of $S$). A vector $V^\mu$ transforms like, $$U^{-1}V^\mu(x)U = {\Lambda^\mu}_\nu V^\nu (\Lambda ^{-1}x),$$ that is for every free Minkowski index, we get an additional ${\Lambda^\mu}_\nu$ on the right-hand side.
$\gamma\textbf{-matrices.}$ We have some matrices $\gamma^\mu$ ($\mu=0,1,2,3$ in 4 spacetime dimensions) with their defining property, $$\{\gamma^\mu, \gamma^\nu\}\equiv \gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2\eta^{\mu\nu}\textbf{1},$$ as well as a fifth $\gamma$-matrix $\gamma^5 := \text{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3 $. We can combine them into five terms with definite index structure: $\Gamma \in \{\textbf{1}, \gamma^\mu, \gamma^5, \gamma^\mu\gamma^5, \gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu\}$ (you may think of additional combinations, but they are not independent of the ones I wrote here, see e.g. here.)
According to the indices, we can assign {scalar, vector, scalar, vector, tensor} ("tensor" is just a name for objects with more indices than a vector). Almost finished.
$\textbf{Parity transformation.}$ Parity acts on quantum fields like $\gamma^0$. It doesn't change the scalar part $\textbf 1$, but since $\gamma^0$ anti-commutes with every other $\gamma$ matrix, it also anti-commutes with $\gamma^5$. So we pick up a minus sign if parity acts on $\gamma^5$, $$\gamma^0 (\gamma^5...) = -\gamma^5\gamma^0...,$$ this means objects built with $\gamma^5$ are no regular scalars, they are $\textit{pseudo}$scalars.
The same goes for the difference in vectors and axial-vectors (That's why there is a minus sign in the table you posted in the P-column: vectors change sign under party, whereas axial-vectors don't).
$\textbf{Dirac bilinears.}$ Caveat: $\gamma^\mu$ itself is no vector. Only if we sandwich a $\gamma$-matrix in-between two spinors, we get something that transforms properly under Lorentz transformations. Finally,
$$
\begin{array}{ll}\hline
\text{Dirac bilinear} & \text{transformation properties}\\\hline
\bar \psi\psi & \text{scalar}\\
\bar \psi\gamma^5\psi & \text{pseudoscalar}\\
\bar \psi\gamma^\mu\psi & \text{vector}\\
\bar \psi\gamma^\mu\gamma^5\psi & \text{axial-vector}\\
\bar \psi(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu)\psi & \text{tensor}\\\hline
\end{array}
$$
$\textbf{Examples.}$
$$
\begin{array}{lllll}\hline
\text{meson} & \text{spin (# of Lorentz indices)} & \text{parity} & J^{P} & \text{Dirac structure}\\\hline
\pi^+ & 0 & - & 0^- & \gamma^5\\
\omega^\mu & 1 & - & 1^- & \gamma^\mu\\
a_0^- & 0 & + & 0^+ & \textbf{1}\\\hline
\end{array}
$$
This post imported from StackExchange Physics at 2020-10-29 11:43 (UTC), posted by SE-user Stephan
Q&A (4870)
