Our starting point is Eq.\[eq:cde\] which we shall write with a simplified notation for two channels. Until the last step in the argument, we shall assume we are using units in which $\hbar^2/2m =1$.

The radial wavefunctions in channel $i$ with incoming waves in channel $\lambda$ are written as $\psi_{i\lambda}$ and have asymptotic behaviour at $r\to\infty$: $$\psi_{i\lambda} \rightarrow \delta_{i\lambda}I_{\lambda} - S_{\lambda i}
O_{i}$$ where for simplicity we write $O_i$ for the outgoing Coulomb wavefunction with orbital angular momentum $\ell$ appropriate to channel $i$, and similarly for the ingoing wavefunction $I_{i}$. For brevity we omit labels for conserved quantum numbers $J$ and $\pi$.

Absorbing the centrifugal interaction within the potential, we can write the coupled equations for the radial wavefunctions appropriate to incoming waves in channel $\lambda$ as:

$$\psi_{i\lambda}'' = 
\sum_j (V_{ij} -E\delta_{ij})\psi_{j\lambda}, \qquad i=1,2.
\label{eq:one}$$

In the case considered here, $V_{ij}$ with $i\ne j$ arises entirely from the tensor interaction, the matrix $V_{ij}$ being symmetric. Now, denoting by $\bar{\psi}$ the wavefunction arising from a (symmetric) perturbation in the potential $V_{ij}\rightarrow V_{ij} + \Delta V_{ij}$, we can write: $$\bar{\psi}_{i\lambda}'' = \sum_j (V_{ij} +\Delta V_{ij} 
-E\delta_{ij})\bar{\psi}_{j\lambda}, \qquad i=1,2.
 \label{eq:two}$$ Multiplying Eq.\[eq:one\] by $\bar{\psi}_{i\mu}$ and Eq.\[eq:two\] by $\psi_{i\mu}$, summing over $i$ and then 