Understanding the Spin States of Vector Mesons

[dwight ang]

Hi All
Thanks for welcoming me to the physics forum. I am reading Greiner and Mueller's Quantum Mechanics: Symmetries and am stuck at not understanding the vector meson ( rho meson)'s spin states.
For S=1 we get three states -1, 0, and 1.
Prof. Mueller separated them as a rho +/-1, and rho 0 .
for the rho +/-1 he mentions the state u ( spin up , spin down) *d_ (read "d bar" for anti d) ( spin up , spin down)
How did he get to this? I did the Clebsch Gordan analysis and got lost . Need some help here.
Thanks
Dwight

 

[vanhees71]

You must be careful to distinguish spin and isospin properly. The $\rho$ meson has spin 1 and isospin 1, i.e., there are three spin states for each charge state, i.e., all together you have 9 physical field-degrees of freedom. In the SU(2) model (only up and down quarks) the electric charge is given by the eigen vectors of $\hat{\tau}_3$, which are $t_3 \in \{-1,0,1 \}$.

Sometimes it's more convenient to work in the SO(3) representation of the isospin. Then you have three real vector fields, written as $\vec{\rho}^{\mu}$. You can easily convert from one to the other isospin basis:
$$\rho^{(\pm) \mu}=\frac{1}{\sqrt{2}} (\rho_1^{\mu} \pm \mathrm{i} \rho_2^{\mu}), \quad \rho^{(0) \mu}=\rho_3.$$

In terms of quark currents the $\rho$ mesons are built from the vector-isovector currents
$$\vec{j}^{\mu}=\overline{\psi} \vec{\tau} \gamma^{\mu} \psi,$$
where $\psi$ is the isospin doublet
$$\psi=\begin{pmatrix} u \\ d \end{pmatrix}$$
and the $\vec{\tau}$ are represented by $\vec{\tau}=\vec{\sigma}/2$, where $\vec{\sigma}$ are the usual Pauli matrices.

By identifying the $\vec{\rho}$ with these currents, you can easily read off the quark content of the $\rho$ mesons. Of course the $\rho^{+}$ is a $u \bar{d}$, the $\rho^{-}$ a $d \bar{u}$, and the $\rho_0$ is given by $\rho_3 \sim |u \bar{u} \rangle - |d \bar{d} \rangle$, because $\tau_3=\mathrm{diag}(1,-1)$.