that the projector on $|\phi_{\alpha}^{JM \pi}\rangle$ in Eq. (\[Bloch\]) is not essential in a microscopic calculation and can be dropped since the various channels are orthogonal at the channel radius.

The Bloch operator ensures that the logarithmic derivative of the wave function is continuous at the channel radius. In addition, $\Psi_{\rm int}^{JM\pi}$ must be equal to $\Psi_{\rm ext}^{JM\pi}$ at $\rho_\alpha=a$. Projecting the Schrödinger equation on a basis state, one obtains $$\begin{aligned}
\sum_{\alpha n} C_{\alpha' n', \alpha n} \, f_{\alpha n} 
= \langle \Phi_{\alpha' n'}^{JM\pi} \vert {\cal L} \vert \Psi_{\rm ext}^{JM\pi} \rangle
\label{SEq}\end{aligned}$$ with $$\begin{aligned}
C_{\alpha' n', \alpha n} = 
\langle \Phi_{\alpha' n'}^{JM\pi} \vert H+{\cal L} -E 
\vert {\cal A}\Phi_{\alpha n}^{JM\pi} \rangle_{\rm int},\end{aligned}$$ and $$\Phi_{\alpha n}^{JM\pi} = u_{\alpha n}(\rho_\alpha)\phi_{\alpha}^{JM\pi}.$$ Here $\langle\vert {\cal O} \vert\rangle_{\rm int}$ indicates that the integration with respect to $\rho_\alpha$ is to be carried out in the internal region. Actually $\langle\vert {\cal O} \vert\rangle_{\rm int}$ is obtained by calculating the matrix element $\langle\vert {\cal O} \vert\rangle$ in the entire space and subtracting the corresponding external matrix element $\langle\vert {\cal O} \vert\rangle_{\rm ext}$ that is easily obtained because no intercluster antisymmetrization is needed. The $R$-matrix and $Z$-matrix are defined by $$\begin{aligned}
&&{\cal R}_{\alpha' \alpha} \equiv \frac{\hbar^2 a}{2}
\left(\frac{k_{\alpha'}}{\mu_{\alpha'}\mu_{\alpha}k_\alpha}\right)^\frac{1}{2}
\sum_{n' 