Two quarks can bound to form a meson, and the wavefunction consists of several quantum degrees of freedom:

$$ \psi=\psi(\text{flavor})\psi(\text{color})\psi(\text{orbital})\psi(\text{spin}) $$

According to Wikipedia, there are several possibilities of mesons: enter image description here

How could we write the representative wavefunction form for all of them? e.g. Pseudoscalar meson, Pseudovector meson, Vector meson, Scalar meson, Tensor meson. In particular, how would the Tensor meson differs from other ones ((Pseudo)-scalar/vector meson)?

(We can assume just $u,d,s$ quarks, and $r,g,b$ colors.)

  • $\begingroup$ The $S$ and $L$ columns in the included table don't look right to me (incomplete rather than incorrect). I changed them on Wikipedia this morning. Last night someone decided that you can make a $0^+$ state from $S=0$ and $L=1$, $i.e$ $|0 - 1|\leq 0 \leq 0 + 1$. I'm not sure if that was vandalism or not... $\endgroup$
    – dukwon
    Commented Sep 25, 2017 at 15:25
  • $\begingroup$ @dukwon, not sure what do you mean, there is no $S=0, L=1$ state in $J^P=0^+$ state on Wiki page? $\endgroup$ Commented Oct 1, 2017 at 3:14
  • $\begingroup$ It's not there any more because I fixed it. It was introduced in this change: en.wikipedia.org/w/… $\endgroup$
    – dukwon
    Commented Oct 1, 2017 at 10:19

2 Answers 2


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,

$$ \begin{array}{ll}\hline \text{Dirac bilinear} & \text{transformation properties}\\\hline \bar \psi\psi & \text{scalar}\\ \bar \psi\gamma^5\psi & \text{pseudoscalar}\\ \bar \psi\gamma^\mu\psi & \text{vector}\\ \bar \psi\gamma^\mu\gamma^5\psi & \text{axial-vector}\\ \bar \psi(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu)\psi & \text{tensor}\\\hline \end{array} $$


$$ \begin{array}{lllll}\hline \text{meson} & \text{spin (# of Lorentz indices)} & \text{parity} & J^{P} & \text{Dirac structure}\\\hline \pi^+ & 0 & - & 0^- & \gamma^5\\ \omega^\mu & 1 & - & 1^- & \gamma^\mu\\ a_0^- & 0 & + & 0^+ & \textbf{1}\\\hline \end{array} $$

  • $\begingroup$ +1, Thanks so much. For the parity part $P$ is fine, but how is the choice of $S$ ad $P$? I know the whole wavefunction should be anti-symmetriezed. But it will be nice to give a few examples outlining the properties and differences for the 5 cases. $\endgroup$ Commented Sep 27, 2017 at 15:16
  • $\begingroup$ I've edited the post with some examples - did this answer your question? If not, please reformulate your question. $\endgroup$
    – ersbygre1
    Commented Sep 28, 2017 at 0:41
  • $\begingroup$ Thanks, but I am also interested in knowing the wavefunction. Other then the color needs to be a singlet, what else constraint do we have for the spin sectors? $\endgroup$ Commented Oct 1, 2017 at 1:29

You can simply use the idea that any meson corresponds to some representation of the $SU(3)$ group (aka eightfold way by Gell-Mann). Then you just need to classify all possible of fundamental and adjoint representations and relate them to the physical particles - mesons, baryons and other. Finally, the only thing you need is to write the wave-function, which is done by using the matrix representation of the $SU(3)$ group.

  • $\begingroup$ Thanks! But it will be nice to write a few examples. Because it is not obvious are you talking about pseudo ones or not. It is clearly related to Representations [but it is difficult to do than just state in principle.] $\endgroup$ Commented Sep 26, 2017 at 3:54

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