Angular distributions for the PWA study

Curtis A. Meyer

June 8, 2000

Introduction

In this document, we want to look at the photo-produced $a_{2}(1320)$, and examine the effects of photon polarization on various observables. We start with the assumption that the $a_{2}$ is produced via $\pi$ exchange. This implies that for the $a_{2}$, the only allowed values of $m_{j}$ are $\pm 1$. We will look at the reaction:

$$a^{+}_{2}\rightarrow \rho^{\circ}\pi^{+} \rightarrow (\pi^{+}\pi^{-})\pi^{+}$$

In the Godfrey-Jackson, (GJ), frame, the $a_{2}$ decays with decay angles $\theta$ and $\phi$. We then move into the $\rho$ helicity frame, where the $\rho$ decays with angles $\theta^{\prime}$ and $\phi^{\prime}$.

Under these assumptions, the most general density matrix for the initial $a_{2}$, $\rho_{\left[a_{2}a_{2}\prime\right]}$ can be written as in 1 in terms of four parameters. There are also two relations between these, $\alpha + \beta = 1$ and $\delta = \gamma^{*}$. We can simplify this by writeing $\eta=\mathrm{Real}(\gamma)$ and $\xi=\mathrm{Imaginary}(\gamma)$.

$$\rho_{\left[a_{2}a_{2}\prime\right]} = \left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & 0 \\ 0 & \alpha & 0 &... ...0 & 0 \\ 0 & \delta & 0 & \beta & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ) = \left( \begin{array}{ccccc} 0 & 0 & 0 & 0 & 0 \\ 0 & \alpha & ... ...& 0 \\ 0 & \eta-i\xi & 0 & \beta & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right)$$ (1)

The density matrix for the $\rho$ can be obtained from the $a_{2}$ via the operation in 2.

$$\rho_{\left[\rho\rho\prime\right]} = \sum_{a_{2},a_{2}\prime... ...t]} f^{\dagger}_{\left[a_{2}a_{2}\prime;\rho\rho\prime\right]}$$ (2)

The $a_{2}$ decays to $\rho\pi$ with relative orbital angular momentum of $L=2$, which means that the transition amplitudes, $f$ depend on the $T$'s in equation 3.

$$T_{\left[\lambda_{1}\lambda_{2}\right]} = \langle J \lambda... ...le S \lambda \mid S_{1} S_{2} \lambda_{1} -\lambda_{2} \rangle$$ (3)

There are three possible non-zero terms, of which two turn out to be non-zero. These are given below.

$$\begin{eqnarray*} T_{\left[ 10\right ]} &= \langle 2 1 \mid 2 1 0 1 \rangle \l... ...rangle \langle 1 -1 \mid 1 0 -1 0 \rangle &= \frac{1}{\sqrt{2}} \end{eqnarray*}$$

The transition amplitude, $f$ can then be written as in equation 4. We have simplified $f$ by explicitly setting to zero all terms that cannot contribute the $\rho_{\left[\rho\rho\prime\right]}$. In particular, $T_{\left[00\right]}$ removes all of the $d^{2}_{m0}$ terms, and the density matrix of the $a_{2}$ means that only the $d^{2}_{\pm 1\pm 1}$ terms will remain.

$$f_{\left[ a_{2}a_{2}\prime;\rho\rho\prime\right]} = \left( \begin{array}{ccccc} 0 & -\frac{1}{\sqrt{2}}d^{2}_{11}(\th... ... & 0 & \frac{1}{\sqrt{2}}d^{2}_{-1-1}(\theta)e^{-i\phi} & 0 \end{array}\right )$$ (4)

The $d$ functions are given as:

$$\begin{eqnarray*} d^{2}_{11}(\theta) &= d^{2}_{-1-1}(\theta) &= \frac{1}{2}(1+\... ...heta) &= \frac{1}{2}(1-\cos\theta)(2\cos\theta + 1) == B(\theta) \end{eqnarray*}$$

We can now use equation 2 to find that the density matrix for the $\rho$. We take $\eta=\mathrm{Real}(\gamma)$ and $\xi=\mathrm{Imaginary}(\gamma)$, and find that $\rho_{\left[\rho\rho\prime\right]}$ is given as:

$$\begin{eqnarray*} \left ( \begin{array}{ccc} \frac{1}{2}(\alpha A^{2}+\beta B^{2... ...B\left[\eta\cos 2\phi -\xi\sin 2\phi\right] \end{array} \right ) \end{eqnarray*}$$

The angular distribution of the spectator $\pi^{+}$ as seen in the GJ frame is given as the trace of $\rho$. We can extract this and find the angular distribution as is 5. We note that if the $a_{2}$ is unpolarized, then $\alpha=\beta=\frac{1}{2}$ and $\eta=\xi=0$. This then yields the distribution in equation 6 which is independent of $\phi$.

$$w(\theta,\phi) = \frac{1}{2}(A^{2}+B^{2}) + 2AB \left[ [\eta\cos 2\phi -\xi\sin 2\phi\right]$$ (5)

$$w_{\mathrm{unpolarized}}(\theta,\phi) = \frac{1}{2}(A^{2}+B^{2})$$ (6)

We can now continue this by looking at the angular distributions of the $\pi^{+}$ and $\pi^{-}$ in the helicity frame, $w(\theta^{\prime}, \phi^{\prime})$. We can rewrite $\rho_{\left[\rho\rho\prime\right]}$ in terms of three real parameters:

$$\begin{eqnarray*} C(\theta,\phi) &=& \frac{1}{2}\left[\alpha A^{2}(\theta)+\beta... ...B^{2}(\theta)\right] \left[\eta\cos 2\phi -\xi\sin 2\phi\right] \end{eqnarray*}$$

These can then be used with equation 7 to determine the angular distributions in the helicity frame.

$$w(\theta,\phi;\theta^{\prime},\phi^{\prime}) = \sum_{\rho\rh... ...rho\prime\right]} f^{\dagger}_{\left[ \rho\rho\prime \right] }$$ (7)

The transition amplitudes for $\rho\rightarrow\pi\pi$ are given as follows:

$$\begin{eqnarray*} f_{\left[\rho\rho\prime\right]} &=& \left(\begin{array}{ccc} d... ...}_{0-1}(\theta^{\prime})e^{-i\phi^{\prime}} \end{array}\right ) \end{eqnarray*}$$

which when combined with the density matrix, and using the fact that $d^{1}_{01}(\theta^{\prime} = \frac{1}{2}\sin\theta^{\prime}$ and $d^{1}_{0-1}(\theta^{\prime} = -\frac{1}{2}\sin\theta^{\prime}$, we find equation 8.

$$w(\theta,\phi;\theta^{\prime},\phi^{\prime}) = \sin^{2}\theta^{\prime} \left\{ \frac{1}{2}\left[C(\theta,\phi)+D(\theta,\phi)\right] - E(\theta,\phi)\cos 2\phi^{\prime} \right \}$$ (8)

It is interesting to look carefully at this weight. The $\phi^{\prime}$ dependence is given entirely by the $\cos2\phi^{\prime}$ term. Note that the parameter $E$ is non zero in both the case of polarized and unpolarized $a_{2}$. As such, the $\phi^{\prime}$ dependence is independent of the polarization. What is true is that the size of the $C$, $D$ and $E$ terms do depend on the polarization, the the size of the $\cos2\phi^{\prime}$ piece relative to the flat piece can vary with the polarization. Let us also look carefully at $E(\theta,\phi)$.

$$E(\theta,\phi) = \frac{1}{2}(\alpha B^{2}(\theta)+\beta A^{2... ...+ A(\theta)B(\theta)\left[\eta\cos 2\phi -\xi\sin 2\phi\right]$$ (9)

For a given choice of $\alpha$, $\beta$, $\eta$ and $\xi$, there will be values of $\theta$ and $\phi$ which will make $E=0$. If we can choose these then the $\cos2\phi^{\prime}$ dependence will vanish.

However, the bottom line of all of this is that the clearest signal for polarized beam is in the $\phi$ distribution of the $\pi^{+}$ in the GJ frame. For a polarized photon beam, choosing events in which the polarization vector is nearly in the production plane, or nearly normal to the production plane will allow us to single out the $\eta$ and $\xi$ parts of the $\phi$ distribution. In the former, we should see $\cos 2\phi$, while in the latter it should be $\sin 2\phi$.