Amplitudes for the Exotic b1π Decay

Angular Distribution of Two-Body Decay
Let's begin with a general amplitude for the two-body decay of a state with angular momentum quantum numbers J,m. Specifically, we want to know the amplitude of this state for having daughter 1 with momentum direction $$\Omega=(\phi,\theta)$$ in the center of mass reference frame, and helicity $$\lambda_1$$, while daughter 2 has direction $$-\Omega=(\phi+\pi,\pi-\theta)$$ and helicity $$\lambda_2$$.

Let U be the decay operator from the initial state into the given 2-body final state. Intermediate between the at-rest initial state of qn J,m and the final plane-wave state is a basis of outgoing waves describing the outgoing 2-body state in a basis of good J,m and helicities. Insertion of the complete set of intermediate basis vectors, and summing over all intermediate J,m gives

\langle \Omega \lambda_1 \lambda_2 | U | J m \rangle = \langle \Omega \lambda_1 \lambda_2 U | J m \rangle $$
 * J m \lambda_1 \lambda_2 \rangle \langle J m \lambda_1 \lambda_2 |

This is one way to describe the final state, but it is not the only way. Instead of specifying the final-state particles' spin state via their helicities, we can first couple their spins together independent of their momentum direction, to obtain total spin S, then couple S to their relative orbital angular momentum L to obtain their total angular momentum J. When we do this, we give up our knowledge of the particles' helicities, having replaced those two quantum numbers with the alternative pair L,S. These two bases, the helicity basis and the L,S basis, are each individually complete and orthonormal within themselves. Following on from the above expression, let us insert a sum over the L,S basis.

\langle \Omega \lambda_1 \lambda_2 | U | J m \rangle = \sum_{L,S} \langle \Omega \lambda_1 \lambda_2 J m L S \rangle \langle J m L S | U | J m \rangle $$
 * J m \lambda_1 \lambda_2 \rangle \langle J m \lambda_1 \lambda_2 |

=\sum_{L,S} \left[ \sqrt{\frac{2J+1}{4\pi}} D_{m \lambda}^{J *}(\Omega,0) \right] \left[ \sqrt{\frac{2L+1}{2J+1}} \left(\begin{array}{cc|c} L & S      & J \\ 0 & \lambda & \lambda \end{array}\right) \left(\begin{array}{cc|c} S_1      &  S_2       & S \\ \lambda_1 & -\lambda_2 & \lambda \end{array}\right) \right] a_{L S}^{J} $$ where $$\lambda=\lambda_1-\lambda_2$$,and the double-stacked symbols are Clebsh-Gordon coefficients. The product of CG coefficients in the second brackets on the right-hand side represent the overlap between the basis vectors in the helicity and L,S basis, which turns out to be independent of m, as required by rotational invariance. This overlap integral is somewhat lengthy to calculate, but the result turns out to be fairly simple, as shown above. This expression holds regardless of what axis is used to define the quantization direction for m, but whatever choice is made must serve as the z-axis of the reference frame in which the plane wave direction $$\Omega$$ is defined.

Isospin Projections
One must also take into account the various ways that the isospin of the daughters can add up to the isospin quantum numbers of the parent, requiring a term:



C^{a,b} = \left(\begin{array}{cc|c} I^a   &  I^b   &  I \\ I_z^a & I_z^b & I_z^a+I_z^b \end{array}\right) $$

where a=1 and b=2 refer to the daughter index. If the two daughter particles belong to the same isospin multiplet, there is a constraint introduced between orbital angular momentum and total isospin that follows from the symmetry of exchanging the two particle identities, because 180 degree rotation is equivalent to the exchange of the daughter identities (a,b becoming b,a). For example, for a two-pion final state in an even-L angular wave, only even I is allowed, and for an odd-L angular wave, only odd I is allowed. Because of this, it is convenient to define a symmetrized variant of the C coefficients defined above,

C(L)=\frac{1}{\sqrt{2}} \left[ C^{a,b} + (-1)^L C^{b,a} \right] $$ It should be kept in mind that this $$C(L)$$ is only applicable for particle pairs in the same isospin multiplet.

Reflectivity
Beside rotational invariance, parity is also a good symmetry of strong hadron dynamics. In the case described above of the decay of a single particle at rest into two daughters, parity conservation places constraints between different final state amplitudes. Instead of considering the parity operator directly, it is convenient to consider the reflectivity operator R. Reflectivity is the product of parity with a 180 degree rotation about the y axis. The advantage of using this more complicated operator to express the constraints of parity is that a general two-particle plane wave basis can be constructed out of eigenstates of reflectivity, whereas a complete plane-wave basis of parity eigenstates is possible only in the restricted case that daughters 1 and 2 are identical. Regardless of the additional rotation, the basic constraint of reflectivity conservation is nothing more than parity conservation plus rotational invariance.

Acting on a state of good J,m, the reflectivity operator has a particularly simple effect.
 * $$\mathbb{R}| J m \rangle = P(-1)^{J-m} | J \; -m \rangle $$

where P is the intrinsic parity of the system. The eigenstates of the reflectivity operator are formed out of states of good J,m as follows.
 * $$| J m \epsilon \rangle = | J m \rangle + \epsilon P (-1)^{J-m} | J \; -m \rangle   $$

where &epsilon;=&plusmn;1 for a bosonic system and &epsilon;=&plusmn;i for a fermionic system. It follows that
 * $$\mathbb{R}| J m \epsilon \rangle = \epsilon (-1)^{2J} | J m \epsilon \rangle $$

Photon-Reggeon-Resonance vertex
Consider the production of the resonance from the photon and reggeon in the reflectivity basis, the eigenstates of the reflectivity operator.

The photon linear polarization states turn out to be eigenstates of reflectivity as well: Let x (y) polarization states be denoted with - (+)

$$|\mp\rangle = \sqrt{\frac{\pm 1}{2}} \left( |1 -1\rangle \mp |1 +1\rangle \right)$$

$$\mathbb{R}|\mp\rangle = \mp 1 |\mp\rangle $$

Since the production Hamiltonian should commute with reflectivity: $$V=\mathbb{R}^{-1} V \mathbb{R}$$

$$ \langle J m \epsilon|\mathbb{R}^{-1} V \mathbb{R}| \mp ; J_R \lambda_R \epsilon_R ; t, s; \Omega_0 \rangle = \epsilon (\mp 1) \epsilon_R \langle J m \epsilon|V| \mp ; J_R \lambda_R \epsilon_R ; t, s; \Omega_0 \rangle $$

Acting with the reflectivity operator on initial and final state brings out the reflectivity eigenvalues of the resonance, photon and reggeon. This result leads to a constraint: $$\epsilon = \mp \epsilon_R$$

Proton-Reggeon vertex
The amplitude of target proton's emission of an exchange particle, a reggeon, in particular direction and helicity projections can be written as:

Decay
$$ \langle \Omega_X 0 \lambda_{b_1} | U_X | J_X m_X \rangle =\sum_{L_X} \left[ \sqrt{\frac{2J_X+1}{4\pi}} D_{m_X \lambda_{b_1}}^{J_X *}(\Omega_X,0) \right] \left[ \sqrt{\frac{2L_X+1}{2J_X+1}} \left(\begin{array}{cc|c} L_X & 1            & J_X \\ 0   & \lambda_{b_1} & \lambda_{b_1} \end{array}\right) \right] a_{L_X}^{J_X} $$

$$ \langle \Omega_{b_1} 0 \lambda_\omega | U_{b_1} | 1, m_{b_1}=\lambda_{b_1} \rangle =\sum_{L_{b_1}} \left[ \sqrt{\frac{2J_{b_1}+1}{4\pi}} D_{m_{b_1}=\lambda_{b_1} \lambda_\omega}^{1 *}(\Omega_{b_1},0) \right] \left[ \sqrt{\frac{2L_{b_1}+1}{2J_{b_1}+1}} \left(\begin{array}{cc|c} L_{b_1} & 1             & 1 \\ 0       & \lambda_\omega & \lambda_\omega \end{array}\right) \right] b_{L_{b_1}} $$

$$ \langle \Omega_\omega 0 \lambda_\rho | U_\omega | 1, m_\omega=\lambda_\omega \rangle =\sum_{L_\omega J_\rho} \left[ \sqrt{\frac{2J_\omega+1}{4\pi}} D_{m_\omega=\lambda_\omega \lambda_\rho}^{1 *}(\Omega_\omega,0) \right] \left[ \sqrt{\frac{2L_\omega+1}{2J_\omega+1}} \left(\begin{array}{cc|c} L_\omega & 1           & 1 \\ 0        & \lambda_\rho & \lambda_\rho \end{array}\right) \right] c_{L_\omega J_\rho} $$

$$ \langle \Omega_\rho 0 \lambda_\rho | U_\rho | J_\rho, m_\rho=\lambda_\rho \rangle =\sum_{L_\rho} \left[ \sqrt{\frac{2J_\rho+1}{4\pi}} D_{m_\rho 0}^{J_\rho *}(\Omega_\rho,0) \right] \left[ \sqrt{\frac{2L_\rho+1}{2J_\rho+1}} \left(\begin{array}{cc|c} L_\rho & 0 & J_\rho \\ 0     & 0 & 0 \end{array}\right) \right] d_{L_\rho} =\sum_{L_\rho} \sqrt{\frac{2L_\rho+1}{4\pi}} Y_{m_\rho}^{J_\rho *}(\Omega_\rho) d_{L_\rho} $$

$$ A^{J_X}=\sum_{\lambda_{b_1},\lambda_\omega,\lambda_\rho} \langle \Omega_X 0 \lambda_{b_1} | U_X | J_X m_X \rangle k^{L_X} \langle \Omega_{b_1} 0 \lambda_\omega | U_{b_1} | 1, m_{b_1}=\lambda_{b_1} \rangle q^{L_{b_1}} \langle \Omega_\omega 0 \lambda_\rho | U_\omega | 1, m_\omega=\lambda_\omega \rangle u^{L_\omega} \langle \Omega_\rho 0 \lambda_\rho | U_\rho | J_\rho, m_\rho=\lambda_\rho \rangle C_\rho(L_\rho) v^{L_\rho} $$