# Vector meson spin states

1. Nov 3, 2015

### 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

2. Nov 8, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 9, 2015

### 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 $\tau_3=\mathrm{diag}(1,-1)$.