Parity Relations in the Reflectivity Basis

Paul Eugenio
and
Curtis A. Meyer

September 21, 2000

Parity Relations of Production Strengths

We want to consider the photo-production process $\gamma N\rightarrow X N'$ and to derive partity relations between the production strengths, $V$, as seen in the reflectivity basis. These will then be combined with the coresponding decay amplitudes $A$ as cmputed in the reflectivity basis to allow us to compute event weigths for our PWA analysis. We start initially with conservation of parity in the helicity frame. For the process $\gamma N\rightarrow X N'$, we know that the following relation ship holds.

$$\begin{eqnarray*} V_{-\lambda_{x}-\lambda_{N'};-\lambda_{\gamma}-\lambda_{N}} &=... ...right]} V_{\lambda_{x}\lambda_{N'};\lambda_{\gamma}\lambda_{N}} \end{eqnarray*}$$

The $\lambda_{i}$ refer to the helicities of the indicated particles, the $J_{i}$ refer to the spins of the particles, and the $P_{i}$ refer to the parity of the particles. This can be simplified by noting first that three of the parities can be eliminated:

$$P_{\gamma}P_{N}P_{N'} = -1 .$$

Next, we can eliminate a lot of the spin factors as:

$$(-1)^{J_{N'}-J_{\gamma}-J_{N}} = (-1)^{-J_{\gamma}} = -1 .$$

and then we know that the naturality of $X$ is given as:

$$\eta_{x} = P_{x}(-1)^{J_{x}}.$$

Finally, we can also simplify the term involving $\lambda_{i}$. For the case of photo-production we know that $\lambda_{\gamma}=\pm 1$ and also that $\lambda_{x}=\pm 1$. This means that $\lambda_{\gamma}-\lambda_{x} = (-2,0,2)$, and therefore $(-1)^{(\lambda_{\gamma}-\lambda_{x})}=+1$. We can now simplify the parity relationship to:

$$\begin{eqnarray*} V_{-\lambda_{x}-\lambda_{N'};-\lambda_{\gamma}-\lambda_{N}} &=... ...a_{N})} V_{\lambda_{x}\lambda_{N'};\lambda_{\gamma}\lambda_{N}} \end{eqnarray*}$$

For the case of spin-flip , $\lambda_{N'}-\lambda_{N}=\pm1$, while for the case of spin-non-flip , $\lambda_{N'}-\lambda_{N}=0$ which yields:

$$\begin{eqnarray*} V_{-\lambda_{x}-\lambda_{N'};-\lambda_{\gamma}-\lambda_{N}} &=... ...};\lambda_{\gamma}\lambda_{N}} \,\,\,\mathrm{spin\,nonflip} \\ \end{eqnarray*}$$

We can now write these production strengths for the $m_{x}=\lambda_{x}$ states in the reflectivity basis for a produced particle of naturality $\eta_{x}$ in a reflectivity state $\epsilon_{x}$. In this case,

$$^{\epsilon}V_{\lambda_{N}\lambda_{N'}\epsilon_{x}} = \frac{1}... ...\lambda_{\gamma}=-1\lambda_{x}=-\mid m_{x}\mid\right]} \right]$$

Let us now take the case where we only have a spin-flip contributing. In addition, helicity conservation will limit the values of $\lambda_{x}$ to be $\lambda_{\gamma}$. In this case $\lambda_{N'}=-\lambda_{N}$, and we can write that:

$$\begin{eqnarray*} ^{\epsilon}V_{\lambda_{N}-\lambda_{N}\epsilon_{x}^{\pm}} &=& \... ...on_{x} ^{\epsilon}V_{\lambda_{N}-\lambda_{N}\epsilon_{x}^{\pm}} \end{eqnarray*}$$

For the case of only spin-non-flip contributions, we have that $\lambda_{N'}=\lambda_{N}$, and we can write that:

$$\begin{eqnarray*} ^{\epsilon}V_{\lambda_{N}\lambda_{N}\epsilon_{x}^{\pm}} &=& \f... ...lon_{x} ^{\epsilon}V_{\lambda_{N}\lambda_{N}\epsilon_{x}^{\pm}} \end{eqnarray*}$$

The Photon Spin-Density Matrix

The last part is to rotate the spin-density matrix of the photon from the helicity frame, $\rho_{\lambda_{\gamma}\lambda_{\gamma}'}$ to the reflectivity basis $\rho_{\epsilon_{\gamma}\epsilon_{\gamma}'}$. Where $\rho_{\lambda_{\gamma}\lambda_{\gamma}'}= \mid\lambda_{\gamma}> <\lambda_{\gamma}'\mid$. To do this, we recall that the basis states can be transformed as:

$$\begin{eqnarray*} <\epsilon_{\gamma}\mid &=& \frac{1}{\sqrt{2}}\left[ <\lambda=+... ...ft[ <\lambda=+1\mid - \epsilon_{\gamma} <\lambda=-1\mid \right] \end{eqnarray*}$$

$$\begin{eqnarray*} <\epsilon_{\gamma}=+1\mid &=& \frac{1}{\sqrt{2}}\left[ <\lambd... ...c{1}{\sqrt{2}}\left[ <\lambda=+1\mid + <\lambda=-1\mid \right] . \end{eqnarray*}$$

There is also a simple relation between the reflectivity states and the linear polarization states, $\mid\epsilon_{\gamma}=+1>=\mid x>$ and $\mid\epsilon_{\gamma}=-1>=-i\mid y>$. We can now multiply out to obtain the four elements of the spin-density matrix in the reflectivity basis.

$$\begin{eqnarray*} \rho_{++} &=& \frac{1}{2}\left[ \rho_{11}-\rho_{1-1}-\rho_{-1... ...1}{2}\left[ \rho_{11}+\rho_{1-1}+\rho_{-11}+\rho_{-1-1} \right] \end{eqnarray*}$$

We also know that $\rho_{11}+\rho_{-1-1}=+1$ and that $\rho_{-1+1}=\rho_{+1-1}$. Using this, we can simplify the spin-density matrix as follows:

$$\begin{eqnarray*} \rho_{\epsilon_{\gamma}\epsilon_{\gamma}'} &=& \left ( \begin{... ...right] & \left[\frac{1}{2}+\rho_{1-1}\right] \end{array}\right) \end{eqnarray*}$$

For the case of an unpolarized photon, we know that $\rho_{11}=\rho_{-1-1}=\frac{1}{2}$ and that $\rho_{1-1}=\rho_{-11}=0$. This means that the spin-density matrix is as expected:

$$\begin{eqnarray*} \rho_{\epsilon_{\gamma}\epsilon_{\gamma}'} &=& \left ( \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right). \end{eqnarray*}$$

For the case of linearly polarized photons, we know that $\rho_{1-1}=\frac{1}{2}\cos 2\alpha$ where $\alpha$ is the angle between the polarization vector and the normal to the production plane as seen in the Gottfried Jackson frame. In addition, $\rho_{11}=\frac{1}{2}(1+\sin 2\alpha)$ and $\rho_{-1-1}=\frac{1}{2}(1-\sin 2\alpha)$. This allows us to simplify the spin-density matrix to:

$$\begin{eqnarray*} \rho_{\epsilon_{\gamma}\epsilon_{\gamma}'} &=& \left ( \begin{... ...\\ \frac{1}{2}\sin 2\alpha & \cos^{2}\alpha \end{array}\right). \end{eqnarray*}$$

Specific Examples

Now, for the case of $a_{2}$ production via $\pi$ exchange, there are four complex production amplitudes, $V_{\left[\lambda_{N}-\lambda_{N}\epsilon_{x}\right]}$. There are two possible spin-flips for each of the two reflectivity. However, parity will reduce this down to two complex production strengths, $V_{\left[ -1,1,+\right]}= V_{\left[ 1,-1,+\right]}$ and $V_{\left[ -1,1,-\right]}= -V_{\left[ 1,-1,-\right]}$. If we write $A_{\mid m\mid =1}^{+}\left[a_{2}\rightarrow\rho\pi\right]$ and $A_{\mid m\mid =1}^{-}\left[a_{2}\rightarrow\rho\pi\right]$ as the decay amplitudes for the two reflectivity states of the $a_{2}$, then the weight for a particular event is given as:

$$\begin{eqnarray*} w &=& 2 \left\{ \rho_{++} \mid V_{\left[ -1,1,+\right]} A_{\m... ...mid m\mid =1}^{+}\left[a_{2}\rightarrow\rho\pi\right]) \right \} \end{eqnarray*}$$

Using the density matrix from above and noting the fact that $a^{*}b+b^{*}a = 2\mathrm{Re}(ab)$, we can somewhat simplify the above expression to be proportional to:

$$\begin{eqnarray*} w &=& \sin^{2}\alpha \mid R^{+} \mid^{2} + \cos^{2}\alpha \mid R^{-} \mid^{2} + \sin\alpha\cos\alpha\mathrm{Re}(R^{+}R^{-}) \end{eqnarray*}$$

Where $R^{\pm}$ is the product of $V^{\pm}$ times $A^{\pm}$.