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Amplitudes for the Exotic b1π Decay (view source)
Revision as of 22:20, 15 August 2011
, 22:20, 15 August 2011no edit summary
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 quantum numbers (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
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 quantum numbers (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
:<math>
:<math>
\langle \Omega \lambda_1 \lambda_2 | U | J m \rangle
\langle \Omega \lambda_1 \lambda_2 | U | J M \rangle
=
=
\langle \Omega \lambda_1 \lambda_2
\langle \Omega \lambda_1 \lambda_2
| J m \lambda_1 \lambda_2 \rangle \langle J m \lambda_1 \lambda_2 |
| J M \lambda_1 \lambda_2 \rangle \langle J M \lambda_1 \lambda_2 |
U | J m \rangle
U | J M \rangle
</math>
</math>
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.
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.
:<math>
:<math>
\langle \Omega \lambda_1 \lambda_2 | U | J m \rangle
\langle \Omega \lambda_1 \lambda_2 | U | J M \rangle
=
=
\sum_{L,S}
\sum_{L,S}
\langle \Omega \lambda_1 \lambda_2
\langle \Omega \lambda_1 \lambda_2
| J m \lambda_1 \lambda_2 \rangle \langle J m \lambda_1 \lambda_2 |
| J M \lambda_1 \lambda_2 \rangle \langle J M \lambda_1 \lambda_2 |
J m L S \rangle \langle J m L S |
J M L S \rangle \langle J M L S |
U | J m \rangle
U | J M \rangle
</math>
</math>
:<math>
:<math>
Acting on a state of good ''J,m'', the reflectivity operator has a particularly simple effect.
Acting on a state of good ''J,m'', the reflectivity operator has a particularly simple effect.
:<math>\mathbb{R}| J m \rangle = P(-1)^{J-m} | J \; -m \rangle </math>
:<math>\mathbb{R}| J M \rangle = P(-1)^{J-m} | J \; -M \rangle </math>
where P is the intrinsic parity of the system.
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.
The eigenstates of the reflectivity operator are formed out of states of good ''J,M'' as follows.
:<math>| J m \epsilon \rangle = | J m \rangle + \epsilon P (-1)^{J-m} | J \; -m \rangle </math>
:<math>| J M \epsilon \rangle = | J M \rangle + \epsilon P (-1)^{J-M} | J \; -M \rangle </math>
where ε=±1 for a bosonic system and ε=±i for a fermionic system. It follows that
where ε=±1 for a bosonic system and ε=±i for a fermionic system. It follows that
:<math>\mathbb{R}| J m \epsilon \rangle = \epsilon (-1)^{2J} | J m \epsilon \rangle </math>
:<math>\mathbb{R}| J M \epsilon \rangle = \epsilon (-1)^{2J} | J M \epsilon \rangle </math>
= Applications =
= Applications =
:<math>
:<math>
\langle J m \epsilon|\mathbb{R}^{-1} V \mathbb{R}|
\langle J M \epsilon|\mathbb{R}^{-1} V \mathbb{R}|
\epsilon_\gamma ; J_R \lambda_R \epsilon_R ; t, s; \Omega_0 \rangle =
\epsilon_\gamma ; J_R \lambda_R \epsilon_R ; t, s; \Omega_0 \rangle =
\epsilon \epsilon_\gamma \epsilon_R \langle J m \epsilon|V|
\epsilon \epsilon_\gamma \epsilon_R \langle J M \epsilon|V|
\epsilon_\gamma ; J_R \lambda_R \epsilon_R ; t, s; \Omega_0 \rangle
\epsilon_\gamma ; J_R \lambda_R \epsilon_R ; t, s; \Omega_0 \rangle
</math>
</math>
<math>
<math>
\langle \Omega_{b_1} \lambda_{b_1} 0| U_X | J_X m_X \rangle
\langle \Omega_{b_1} \lambda_{b_1} 0| U_X | J_X M_X \rangle
=\sum_{L_X}
=\sum_{L_X}
\left[ \sqrt{\frac{2J_X+1}{4\pi}} D_{m_X \lambda_{b_1}}^{J_X *}(\Omega_{b_1}) \right]
\left[ \sqrt{\frac{2J_X+1}{4\pi}} D_{m_X \lambda_{b_1}}^{J_X *}(\Omega_{b_1}) \right]
<math>
<math>
\langle \Omega_\omega \lambda_\omega 0| U_{b_1} | 1 , m_{b_1}=\lambda_{b_1} \rangle
\langle \Omega_\omega \lambda_\omega 0| U_{b_1} | 1 , M_{b_1}=\lambda_{b_1} \rangle
=\sum_{L_{b_1}}
=\sum_{L_{b_1}}
\left[ \sqrt{\frac{2J_{b_1}+1}{4\pi}} D_{m_{b_1} \lambda_\omega}^{1 *}(\Omega_\omega) \right]
\left[ \sqrt{\frac{2J_{b_1}+1}{4\pi}} D_{m_{b_1} \lambda_\omega}^{1 *}(\Omega_\omega) \right]
<math>
<math>
\langle \Omega_\rho \lambda_\rho 0| U_\omega | 1 , m_\omega=\lambda_\omega \rangle
\langle \Omega_\rho \lambda_\rho 0| U_\omega | 1 , M_\omega=\lambda_\omega \rangle
=\sum_{L_\omega J_\rho}
=\sum_{L_\omega J_\rho}
\left[ \sqrt{\frac{2J_\omega+1}{4\pi}} D_{m_\omega \lambda_\rho}^{1 *}(\Omega_\rho) \right]
\left[ \sqrt{\frac{2J_\omega+1}{4\pi}} D_{m_\omega \lambda_\rho}^{1 *}(\Omega_\rho) \right]
<math>
<math>
\langle \Omega_{\pi} 0 0 | U_\rho | J_\rho , m_\rho=\lambda_\rho \rangle
\langle \Omega_{\pi} 0 0 | U_\rho | J_\rho , M_\rho=\lambda_\rho \rangle
=\sum_{L_\rho}
=\sum_{L_\rho}
\left[ \sqrt{\frac{2J_\rho+1}{4\pi}} D_{m_\rho 0}^{J_\rho *}(\Omega_{\pi}) \right]
\left[ \sqrt{\frac{2J_\rho+1}{4\pi}} D_{m_\rho 0}^{J_\rho *}(\Omega_{\pi}) \right]
</math>
</math>
::<math>=\sum_{\lambda_R,\lambda_{b_1},\lambda_\omega,\lambda_\rho,m_X}
::<math>=\sum_{\lambda_R,\lambda_{b_1},\lambda_\omega,\lambda_\rho,m_X}
\langle \mathbf{q}_{b1} \lambda_{b_1} 0| U_X | J_X m_X \epsilon_X\rangle
\langle \mathbf{q}_{b1} \lambda_{b_1} 0| U_X | J_X M_X \epsilon_X\rangle
\langle J_X m_X \epsilon_X | V |
\langle J_X M_X \epsilon_X | V |
\epsilon_\gamma; J_R \lambda_R \epsilon_R; s,t \rangle
\epsilon_\gamma; J_R \lambda_R \epsilon_R; s,t \rangle
</math>
</math>
\langle \mathbf{q} \lambda_1 \lambda_2| U | J , m \rangle =
\langle \mathbf{q} \lambda_1 \lambda_2| U | J , M \rangle =
\sum_L
\sum_L
\langle \Omega \lambda_1 \lambda_2| U | J , m \rangle BW(q)
\langle \Omega \lambda_1 \lambda_2| U | J , M \rangle BW(q)
</math>
</math>